SOLID  GEOMETRY 


'BY""    '• 


CLAUDE  IRWIN  PALMER,  A.B. 

ASSOCIATE  PROFESSOR  OF  MATHEMATICS 
ARMOUR   INSTITUTE  OF  TECHNOLOGY 

AND 

DANIEL  POMEROY  TAYLOR,  A.M. 


Edited  by 
GEORGE  WILLIAM   MYERS,   Ph.D. 

PROFESSOR  OF  THE  TEACHING  OF  MATHEMATICS 

SCHOOL  OF    EDUCATION 

THE    UNIVERSITY    OF    CHICAGO 


SCOTT,  FORESMAN  AND  COMPANY 
CHICAGO  NEW  YORK 


Q.A4-S7 


COPYRIGHT  1918 

BY 

SCOTT,  FORESMAN  AND  COMPANY 

EDUCATION  D£PT* 


0tc 


PREFACE 


The  purpose  of  this  book  is  to  present  the  subject  of  solid 
geometry  so  as  to  emphasize  some  of  its  applications  as  well 
as  to  give  a  thorough  training  in  logical  reasoning.  To  accom- 
plish this  the  authors  have  continued  the  same  method  of 
presentation  that  characterizes  their  Plane  Geometry. 

The  aim  of  the  authors  in  the  preparation  of  this  text  has 
been  to  present  the  subject  in  a  logical  manner  that  is  com- 
prehensible to  the  youthful  mind,  and  to  vitalize  the  subject- 
matter —  making  it  both  interesting  and  useful  through  a 
wide  range  of  practical  applications. 

Some  of  the  features  peculiar  to  the  Solid  Geometry  are  out- 
lined in  the  following  paragraphs: 

1.  In  Chapter  VI,  the  theorems  on  the  relations  of  lines  to 
planes  and  of  planes  to  planes  are  carefully  grouped  so  as  to 
make  them  more  easily .  comprehended  by  the  student. 

2.  Distinct  advantage  is  gained  by  combining  certain  related 
subjects,  thus  securing  brevity  and  simplicity  of  treatment. 
Prisms  are  combined  with  cylinders,  pyramids  with  cones, 
and  polyedral  angles  with  spherical  polygons. 

3.  In  the  theorems  regarding  the  areas  and  volumes  of  solids, 
and  requiring  a  use  of  limits,  great  care  is  taken  to  make  the 
treatment  logical  and  still  keep  it  within  the  grasp  of  the  aver- 
age student. 

4.  The  large  number  and  great  variety  of  exercises  are  care- 
fully distributed,  and  range  from  some  quite  elementary  to 
others  of  considerable  difficulty.  This  enables  the  teacher  to 
adapt  the  book  to  any  group  of  students,  whether  in  a  classical 
or  a  technical  high  school.  The  abstract  exercises  give  mental 
training  and  application  of  the  basic  theorems,  while  the  prac- 
tical exercises  are  used  to  correlate  geometric  facts  with  real 
life. 

54  10*85 


iv  PREFACE 


Many  of  the  exercises  involve  an  application  of  arithmetic 
and  algebra  to  geometry. 

In  addition  to  the  four  points  outlined  above,  attention  is 
called  to  the  following  features: 

While  many  theorems  are  proved  in  full,  the  proof  of  others 
is  given  only  in  part.  In  still  others  the  method  of  proof  is 
suggested  only  or  the  work  is  left  entirely  to  the  student. 
Thus  a  middle  course  is  adopted  in  the  use  of  the  suggestive 
method. 

The  Prismatoid  Formula,  one  of  the  most  powerful  theorems 
of  solid  geometry,  is  proved  in  a  simple  direct  manner,  and  is 
followed  by  various  applications.  However,  this  may  be 
omitted  and  in  no  way  interfere  with  the  proofs  of  later  theo- 
rems. 

The  more  important  theorems  are  printed  in  heavy  type. 
This  is  in  accordance  with  the  recommendation  of  the  Com- 
mittee of  Fifteen,  whose  report,  as  well  as  various  other  com- 
mittee reports,  have  been  given  careful  consideration  in  the 
preparation  of  the  entire  work. 

For  convenience,  a  combination  ruler  and  protractor  accom- 
panies each  book. 

Acknowledgment  is  due  to  the  McGraw-Hill  Book  Co.,  Inc., 
for  permission  to  use  exercises  from  Palmer's  Practical  Mathe- 
matics, in  which  are  gathered  numerous  exercises  from  the 
author's  many  years  of  experience  in  teaching  practical  students. 

C.  I.  Palmer. 
Chicago,  September,  1918.  D.  P.  Taylor. 


CONTENTS 

P*GE 

Introduction 273 

Chapter  VI.     Straight  Lines  and  Planes 

Determining  a  Plane 278 

Relative  Positions  of  Lines  and  Planes      280 

Intersecting  Planes : 281 

Lines  and  Planes,  Parallel ■ 283 

Lines  and  Planes,  Perpendicular 288 

Projections 293 

Diedral  Angles ' 295 

Polyedral  Angles 302 

Loci      307 

General  Exercises 309 

Chapter  VII.     Polyedrons,  Prisms,  Cylinders 

Definitions 313 

Prisms  and  Cylinders 315 

Areas  of  Prisms  and  Cylinders 319 

Congruent  and  Equivalent  Solids 324 

Parallelepipeds 326 

Volumes  of  Prisms  and  Cylinders 331 

Similar  Prisms  and  Cylinders      336 

General  Exercises 339 

Chapter  VIII.     Pyramids  and  Cones 

Definitions  and  Theorems 343 

Areas  of  Pyramids  and  Cones 349 

Area  of  Frustum  of  Pyramid  or  Cone 353 

Volumes  of  Pyramids  and  Cones 357 

Similar  Pyramids  and  Cones 365 

General  Exercises 367 

Chapter  IX.     Prism atoids  and  Polyedrons 

Prismatoids 369 

Polyedrons 373 

General  Exercises 379 

v 


vi  CONTENTS 

Chapter  X.     The  Sphere  and  Polyedral  Angles 

Page 

General  Properties 381 

Circles  of  Spheres 384 

Distance  on  a  Sphere 386 

Relative  Positions  of  Spheres 392 

Measurement  of  the  Sphere,  Area 393 

Volume  of  a  Sphere 400 

Spherical  Angles 407 

Spherical  Polygons  and  Polyedral  Angles »;  408 

Polar  Triangles  • 415 

Areas  of  Spherical  Polygons 420 

General  Exercises 424 

Formulas  for  Reference 427 

Syllabus  of  Plane  Geometry 429 

Index 446 


SOLID  GEGME3;RX  :V 
INTRODUCTION 

528.  Solid  geometry  may  be  thought  of  as  an  extension  of 
plane  geometry,  with  which  it  has  much  in  common.  The 
figures  of  plane  geometry  lie  in  a  plane;  while  in  solid  geom- 
etry we  consider  the  properties  of  figures  that  do  not  lie 
entirely  in  one  plane.  Such  figures  are  sometimes  called  three- 
dimensional  figures. 

529.  Space  Ideas.  We  gain  ideas  concerning  space  and 
objects  in  space  through  the  senses  of  touch,  sight,  and  hear- 
ing. Mainly  through  touch  and  sight,  we  determine  the  size 
and  shape  of  an  object. 

530.  Difficulties.  Most  beginners  in  the  study  of  solid 
geometry  have  difficulty  in  visualizing,  or  forming  a  mental 
image,  of  a  three-dimensional  figure  from  a  worded  descrip- 
tion or  from  a  drawing  in  a  plane. 


Even  the  same  picture  may  present  to  the  minds  of  two 
persons  quite  different  images.  For  instance,  in  the  figure, 
one  person  may  see  six  blocks  while  another  will  see  seven. 

273 


274  SOLID  GEOMETRY 

A  little  practice,  however,  in,  visualizing  the  blocks  enables  one 
at  wil;  to  see  six  or  seven  blocks. 

Frequently  a  student  does  not  understand  a  proposition  or  a 
proof  because  he  does  not  visualize  the  figure  properly.  Lines 
may  appear  to  extend  in  a  different  direction  than  was  intended, 
or  a  plane  may  seem  to  lie  in  front  of  the  page  when  it  was 
intended  that  it  should  appear  back  of  the  page. 

Whatever  device  may  be  necessary  for  a  clear  image  should 
be  used;  but  care  should  be  taken  not  to  become  too  dependent 
upon  models.  A  very  important  part  of  the  training  from  the 
study  of  solid  geometry  is  that  it  trains  in  visualizing  three- 
dimensional  figures  from  a  description  or  a  drawing. 

531.  Devices  to  Assist  in  Imaging.  Cardboard  models 
may  be  formed  as  described  in  §  753. 

Planes  can  be  made  to  intersect  by  cutting  two  pieces  of 
cardboard  half  in  two  and  fitting  together  along  the  cuts. 

Cardboard  can  be  used  in  connection  with  sharp  pointed 
wires,  such  as  hatpins.     §  568. 

Hatpins  or  sharp  wires  held  together  by  small  corks  can 
be  used  to  form  open  models.     §  571. 

Various  figures  can  be  formed  with  the  hatpins  and  a  piece 
of  soft  board. 

532.  Representation  of  Solid  Geometry  Figures.  In  solid 
geometry,  the  figures  are  usually  represented  on  paper  or  on 
the  blackboard,  that  is,  in  a  plane.  It  is  well  to  have  some 
idea  of  how  to  make  the  figures  appear  solid,  and  to  "stand 
out"  as  desired.  While  it  is  not  thought  best  to  enter  into  a 
discussion  of  perspective,  a  few  general  suggestions  that  are 
of  assistance  will  be  given. 

The  figures  may  be  supposed  to  be  viewed  from  a  point  a 
little  above,  and  either  directly  in  front  or  a  little  to  the 
right  or  left.  Parts  of  the  figure  that  are  rectangles,  in  gen- 
eral, appear  as  parallelograms  in  the  drawing.  §§  663-5,  671, 
678.     Polygons  appear  as  polygons,  but  reduced  in  one  direc- 


INTRODUCTION 


275 


tion.  §§  660,  749.  Parts  that  are  circles  appear  as  ellipses, 
which  become  narrower  as  the  eye  of  the  observer  is  nearer 
to  the  plane  in  which  the  circle  lies.     §§  689,  692,  864. 

Planes  are  supposed  to  be  opaque,  and  all  lines  covered  by 
them  should  be  left  out,  or,  if  it  is  desired  to  insert  them, 
they  should  be  dashed  lines. 

Two  lines  that  intersect  in  the  drawing  do  not  necessarily 
intersect  in  the  solid  figure.  Lines  are  not  usually  of  the 
same  length  as  they  are  drawn.  The  size  of  an  angle  may  be 
quite  different  from  the  size  in  the  drawing. 

In  the  accompanying  drawings  representing  cubes,  the  one  at  the  left 
appears  to  have  the  face  ABCD  in  front  and  can  hardly  be  imaged  other- 


71       7 


H 


71 


wise.  While  the  one  at  the  right  can  easily  be  seen  as  if  viewed  from 
a  point  a  little  above  and  to  the  right,  and  so  having  the  face  ABCD 
in  front;  or  it  may  be  visualized  as  viewed  from  a  point  a  little  below 
and  to  the  left,  and  so  having  the  face  EFGH  in  front. 

533.    The  Use  of  Plane  Geometry  in  Solid  Geometry.    All 

of  the  plane  geometry  is  at  the  disposal  of  the  student  in 
solid  geometry;  but,  in  applying  the  facts  of  plane  geometry, 
great  care  must  be  taken  not  to  use  them  except  where  they 
will  hold. 

Facts  of  plane  geometry  that  hold  without  reference  to  any 
plane  are  the  axioms  and  most  definitions,  theorems  stating 
that  figures  are  congruent,  and  all  the  theorems  where  it  is 
evident  that  it  does  not  matter  if  the  figures  are  not  in  the 
same  plane. 

Theorems  of  plane  geometry  that  hold  because  the  nature 
of  the  figure  requires  that  it  lie  in  a  plane  are  those  concern- 


276  SOLID  GEOMETRY 


ing  a  triangle,  two  parallel  lines,  a  parallelogram,  a  circle,  and 
all  other  theorems  where  the  conditions  stated  determine  a 
plane  in  which  the  parts  must  lie. 

In  all  other  cases,  the  theorems  of  plane  geometry  can  be 
applied  only  after  all  the  parts  concerned  are  shown  to  lie  in 
one  plane.  Under  these  will  fall  many  theorems  concerning 
parallels,  perpendiculars,  circles,  and  polygons. 

EXERCISES 

1.  Draw  the  figure  of  §  603,  omitting  the  lines  DE,  EF,  and  DF.  Can 
you  visualize  the  figure  as  otherwise  than  in  one  plane? 

2.  Draw  the  figure  of  §  572,  making  the  side  fines  of  plane  P  full.  Com- 
pare with  the  figure  of  the  text. 

3.  Draw  the  figure  of  §  628,  making  all  lines  full  and  of  the  same  thick- 
ness.    Compare  with  the  figure  of  the  text. 

4.  Draw  the  figure  of  §  671,  making  all  lines  of  the  same  thickness. 
Does  your  figure  appear  as  if  viewed  from  below?  If  not,  make  such  lines 
heavy  as  will  make  it  appear  so. 

5.  Draw  a  figure  like  the  one  at  the  right  in  §  722,  but  change  the 
fines  so  that  it  will  appear  as  if  viewed  from  below. 

6.  Draw  a  figure  of  a  sphere  as  in  §  787,  but  viewed  from  a  point  on  a 
level  with  the  center  of  the  sphere.  * 


CHAPTER  VI 

STRAIGHT  LINES  AND   PLANES 

534.  Surface.  A  boundary  of  a  portion  of  space  is  called 
a  surface.  A  surface  may  be  limited  or  unlimited  in  extent. 
It  has  two  dimensions. 

535.  Plane.  A  surface  such  that  a  straight  line  joining 
any  two  of  its  points,  lies  wholly  in  the  surface,  is  called  a 
plane  surface  or,  simply,  a  plane. 

Since  a  straight  line  may  be  unlimited  in  length,  this  defi- 
nition implies  that  a  plane  may  be  unlimited  in  extent,  for 
otherwise  the  straight  line  could  not  lie  wholly  in  the  surface. 

It  also  follows  from  the  definition  that  a  straight  line  can 
intersect  a  plane  in  but  one  point. 


M 


536.  Representing  a  Plane.  In  drawing  a  geometric  figure 
a  portion  of  the  plane  is  conveniently  represented  as  a  rec- 
tangle seen  obliquely.  In  order  to  make  the  plane  appear  to 
"stand  out,"  the  front  edge  is  frequently  drawn  longer  than 
the  back  edge.  Making  the  edges  on  the  front  and  one  end 
heavy  and  shading  help  one  to  image  a  plane  as  it  is  intended 
it  should  appear. 

537.  Reading  a  Plane.  A  plane  may  be  read  by  a  single 
letter,  by  letters  at  opposite  vertices  of  the  rectangle  that 
represents  it,  or  by  the  letters  at  all  the  vertices  of  the  rec- 
tangle. Thus,  in  the  figure,  the  planes  may  be  read  "plane 
M,"  "plane  AC,"  or  "plane  ABCD." 

277 


STRAIGHT  LINES  AND  PLANES 


279 


EXERCISES 


So  that  they  do 
D 


1.  Hold  two  pencils  so  that  they  determine  a  plane, 
not  determine  a  plane. 

2.  In  making  a  kite,  two  straight  sticks  AB  and  CD 
are  fastened  together  at  P.  Show  that  paper  stretched 
over  these  sticks  lies  in  a  plane. 

3.  Show  that  a  rubber  cord  fastened  at  both  ends 
and  stretched  by  grasping  it  at  any  other  point,  lies  in 
one  plane. 

4.  From  a  given  point  lines  are  drawn  to  points  on 
a  given  line;  show  that  all  these  lines  lie  in  the  same  plane. 

5.  What  is  the  locus  of  all  lines  that  pass  through  a  given  point  and 
intersect  a  given  line  not  containing  the  point? 

6.  Show  that,  if  a  rubber  band  is  stretched  by  three  hooks  in  any 
position,  all  parts  of  the  band  he  in  one  plane. 

7.  Do  four  points  necessarily  he  in  a  plane?    When  do  they?     Give 
illustrations  by  using  points  in  the  classroom. 

8.  Why  is  a  tripod  used  to  support  a  camera  or  a  surveyor's  transit? 

9.  Why  do  the  four  legs  of  a  chair  sometimes  not  all  rest  upon  a  floor? 

10.  How  many  planes   are  determined  by  four 
points  not  all  lying  in  the  same  plane? 

11.  Given  five  points  four  of  which  are  in  the  same 
plane.     How  many  planes  do  they  determine? 

12.  How  many  planes  are  determined  by  three 
concurrent  lines  that  do  not  all  He  in  the  same  plane? 

13.  How  does  a  carpenter  determine  whether  a 
floor  is  a  plane? 

14.  Why  does  a  mason  use  a  trowel  with  long  straight  edges  when 
" truing  up"  a  wall? 

15.  How  many  planes  are  determined  by  four  lines  all  meeting  in  a 
point  but  no  three  of  which  he  in  the  same  plane? 

16.  Must  a  triangle  lie  in  a  plane?  Must  a  parallelogram?  Must  a 
trapezoid?     Must  every  quadrilateral? 

17.  Is  the  following  a  complete  definition  for  a  circle:  A  curved  line 
every  point  of  which  is  equally  distant  from  a  fixed  point  called  the  center 
is  a  circle?     Why? 

18.  Through  a  point  in  space  (1)  draw  a  line  parallel  to  a  given  line; 
(2)  draw  a  line  perpendicular  to  a  given  line. 


280  SOLID  GEOMETRY 

RELATIVE  POSITIONS   OF   LINES  AND   PLANES 

545.  Definitions.  Collinear  means  lying  in  the  same  line. 
Coplanar  means  lying  in  the  same  plane. 

546.  Relations  of  Two  Lines.  From  the  study  of  plane 
geometry  and  §§542,  543,  it  follows: 

(1)  That  two  coplanar  lines  may  coincide,  intersect,  or  be 
parallel. 

(2)  That  two  non-coplanar  lines  are  neither  intersecting  nor 
parallel. 

547.  Relations  of  a  Line  and  a  Plane.  A  straight  line  may 
have  three  positions  relative  to  a  plane: 

(1)  It  may  lie  in  the  plane. 

(2)  It  may  intersect  the  plane. 

(3)  It  may  be  parallel  to  the  plane. 

A  line  lies  in  a  plane  if  all  of  its  points  are  in  common  with 
the  plane. 

A  line  intersects  a  plane  if  it  has  only  one  point  in  com- 
mon with  the  plane.  The  point  in  common  is  called  the  foot 
of  the  line. 

A  line  is  parallel  to  a  plane,  and  the  plane  is  parallel  to  the 
line,  if  they  have  no  point  in  common,  that  is,  if  they  do  not 
meet. 

548.  Relations  of  Two  Planes.  Two  planes  may  have 
three  relative  positions: 

(1)  They  may  coincide. 

(2)  They  may  intersect. 

(3)  They  may  be  parallel. 

Two  planes  are  said  to  coincide  if  they  have  the  same 
determining  conditions. 

In  plane  geometry,  the  relative  positions  of  points  and  lines  are  few  and 
so  the  treatment  of  the  subject  is  quite  informal.  In  solid  geometry,  the 
study  is  made  of  points,  lines,  and  planes,  and  their  relative  positions  are 
many.     It  is  advisable  then,  to  make  a  more  formal  study  of  the  subject. 


INTERSECTING  PLANES 


281 


INTERSECTING   PLANES 

549.  Definition.  The  intersection  of  two  surfaces  consists 
of  all  points  common  to  the  two  surfaces,  and  no  other  points; 
that  is,  it  is  the  locus  of  all  points  common  to  the  two  sur- 
faces. 

550.  Axiom.  If  two  planes  have  one  point  in  common,  they 
also  have  another  point  in  common. 

551.  Theorem.  If  two  planes  intersect,  their  intersection  is 
a  straight  line. 


Given  two  intersecting  planes  P  and  Q. 

To  prove  that  their  intersection  is  a  straight  line. 

Proof.  Let  A  and  B  be  two  points  common  to  both 
planes.  .  §  550 

The  straight  line  determined  by  A  and  B  lies  in  the  plane 
P  and  also  in  Q.  §  535 

Then  AB  is  common  to  the  two  planes.  Why? 

Moreover,  no  point  not  in  A  B  is  common  to  the  two  planes 
for  then  they  would  coincide.  §  540 

.*.  the  intersection  of  planes  P  and  Q  is  a  straight  line. 

EXERCISES 

1.  Can  three  planes  intersect  (1)  in  one  line?  (2)  in  two  lines?  (3)  in 
ohree  lines?  (4)  in  more  than  three  lines?     Explain. 

2.  If  a  paper  is  folded  and  creased,  why  does  it  form  a  straightedge 
ilong  the  crease? 

3.  From  a  point  external  to  two  non-coplanar  lines  only  one  line  can 
)e  drawn  that  will  cut  the  two  lines. 


282  SOLID  GEOMETRY 

552.  Theorem.  If  three  planes,  not  passing  through  the  sam> 
line ,  intersect  each  other,  their  three  lines  of  intersection  either 
meet  in  a  point  or  are  parallel  each  to  each. 


/  '    / 

/  '  / 

Given  the  three  planes  AD,  CF,  and  AF  not  passing  througlj 
the  same  line,  and  intersecting  in  lines  AB,  CD,  and  EF. 

To  prove  that  AB,  CD,  and  EF  either  meet  in  a  point  or  ard 
parallel  each  to  each. 

Proof.  Consider  the  two  lines  of  intersection  lying  in  anj 
plane.     (1)  These  lines  intersect,  or  (2)  they  are  parallel. 

(1)  Suppose  AB  and  CD  intersect  at  P. 

Then  P  lies  in  plane  AF  and  in  plane  CF,  and  hence  in  theij 
intersection  EF.  §  54fl 

.-.  AB,  CD,  and  EF  meet  in  point  P. 

(2)  Suppose  AB  II  CD. 
Then  EF  II  CD  or  intersects  CD. 

Suppose  EF  intersects  CD,  then  AB  intersects  CD  by  (1) 
But  this  is  impossible,  for  AB  II  CD. 
Hence  EF  II  CD. 

Similarly  it  can  be  proved  that  EF  II  AB. 

.'.  AB,  CD,  and  EF  are  parallel  each  to  each. 

EXERCISES 

1.  Three  planes  that  do  not  contain  the  same  straight  line  can  hava 
but  one  point  in  common. 

2.  Explain  the  meaning  of  the  expression,  "Two  planes  determine  a 
straight  line."  Is  it  always  true  that  two  planes  determine  a  straight 
line?  Is  the  analogous  statement,  "  Two  lines  determine  a  point, "  of  plan^ 
geometry  always  true?    Explain. 


0 

'F 


PARALLEL  LINES  AND  PLANES  283 

LINES   AND   PLANES,   PARALLEL 

553.  Theorem.  Two  straight  lines  that  are  parallel  to  a  third 
straight  line  are  parallel  to  each  other.  A  R 

Given  lines  A B  and  CD,  each  parallel 
to  EF. 
To  prove  AB  II  CD. 
Proof.     Either  AB  is  parallel  to  CD  or  it  is  not. 

AB  and  EF  determine  plane  AF.  Why? 

CD  and  EF  determine  plane  CF.  Why? 

AB  and  C  determine  plane  BC.  Why? 

If  AB  is  not  parallel  to  CD,  plane  BC  will  intersect  plane  CF 

in  some  line  CG  other  than  CD.  §  551 

Then  CG  II  EF.  §  552  (2) 

But                                    CD  II  EF.  Given 

Hence                        CD  and  CG  coincide.  §  118 

.-.  AB  II  CD.  Why? 

554.  Theorem.  A  plane  containing  one,  and  but  one,  of  two 
parallel  lines  is  parallel  to  the  other. 

Given  two  parallel  lines  AB  and  CD, 
and  plane  P  containing  CD  but  not  AB. 

To  prove  plane  P  II  AB. 

Suggestion.  AB  and  CD  determine  a  plane.  Suppose  AB 
not  parallel  to  plane  P,  and  show  that  then  AB  would  not  be 
parallel  to  CD. 

EXERCISES 

1.  As  a  door  is  opened,  show  that  all  the  positions  of  its  outer  edge  are 
parallel  to  each  other. 

2.  Prove  that  the  lines  connecting  in  order  the  middle  points  of  the 
sides  of  a  quadrilateral  in  space  form  a  parallelogram. 

By  a  " quadrilateral  in  space"  is  meant  a  quadrilateral  not  all  of  whose 
sides  lie  in  the  same  plane.     It  is  often  called  a  skew  or  gauche  quadrilateral. 


D  p> 


A 

B 

k 

Q 

■■1 

A 

~Q         — — 

B 

t 

0 

-/ 

284  SOLID  GEOMETRY 

555.  Theorem.  If  a  line  is  parallel  to  a 
plane,  it  is  parallel  to  the  intersection  of  that 
plane  with  any  plane  through  the  line. 

556.  Theorem.  A  line  is  in  a  plane  if  it 
has  a  point  in  the  plane,  and  the  line  and  the 
plane  are  both  parallel  to  a  second  line. 

The  two  parallel  lines  determine  a  plane  intersecting  the  first  plane. 

557.  Theorem.  If  a  line  is  parallel  to  the  intersection  of  two 
planes,  it  is  parallel  to  each  of  the  planes. 

558.  Theorem.  A  line  parallel  to  each  of  two  intersecting 
planes  is  parallel  to  their  intersection.  a b 

A  line  through  a  point  in  the  intersection  of  the 
two  planes  and  parallel  to  the  given  line  must  lie  in 
both  planes  by  §  556. 

559.  Problem.  Through  a  given  line  to  pass  a  plane  parallel 
to  any  other  straight  line  not  intersecting  the  given  line. 

(1)  How  many  planes  can  be  passed  through  the  given  line  and  parallel 
to  the  other  line  if  the  given  line  is  parallel  to  the  other  line?     §§  541,  554. 

(2)  If  the  given  line  is  not  parallel  to  the  other,  construct  a  fine  parallel 
to  the  given  fine  and  through  a  point  in  the  other.  Then  the  intersecting 
lines  determine  a  plane  parallel  to  the  given  line. 

560.  Problem.  Through  a  point  to  pass  a  plane  parallel  to 
each  of  two  given  non-coplanar  lines. 

Given  AB  and  CD,  the  two  lines,  and  P, 
the  point. 

Construction.     Through  P  construct  lines 
parallel  to  both  AB  and  CD.     These  lines  determine  the  plane 
parallel  to  A B  and  CD. 

Give  proof. 

EXERCISES 

1.  If  a  line  in  one  of  two  intersecting  planes  is  parallel  to  the  other 
plane,  it  is  parallel  to  their  intersection. 

2.  If  a  line  and  a  plane  are  parallel  to  the  same  line,  they  are  parallel 
to  each  other,  or  the  line  lies  in  the  plane. 


A- 

C\ 

B 

l^G 

L 

^H 

1 

PARALLEL  LINES  AND  PLANES  285 

561.  Definition.  Two  planes  are  parallel  if  they  have  no 
points  in  common,  that  is,  if  they  do  not  meet  however  far 
they  may  be  produced. 

562.  Theorem.  If  a  plane  intersects  two  parallel  planes,  the 
lines  of  intersection  are  parallel. 

Given  plane  PQ  intersecting  the  two  parallel 
planes  MN  and  RS  in  AB  and  CD,  respectively. 

To  prove  AB  II  CD. 

Suggestion.  Since  AB  and  CD  are  in  the 
same  plane  PQ,  they  are  parallel  or  intersect. 
Suppose    that   they   intersect.     What   follows? 

563.  Theorem.  Parallel  lines  intercepted 
between  parallel  planes  are  equal. 

564.  Theorem.  A  line  is  in  a  plane  if  it  has  a  point  in  the 
plane  and  if  the  line  and  plane  are  both  parallel  to  a  second  plane. 

Pass  a  plane  through  the  line  and  intersecting  the  two  planes.  Then 
apply  §  118. 

EXERCISES 

1.  If  a  line  and  a  plane  are  parallel  to  the  same  plane,  they  are  parallel 
to  each  other,  or  the  line  lies  in  the  plane. 

2.  State  the  converse  of  the  theorem  of  §  562.     Is  it  true? 

3.  A  board  is  cut  across  on  a  slant.  What  is  the  shape  of  the  cut 
section? 

565.  Theorem.  If  two  intersecting  straight  lines  are  parallel 
to  a  plane,  the  plane  determined  by  these  lines  is  parallel  to  that 
plane. 

Given  the  intersecting  lines  AB  and  CD,  each  parallel  to  plane 
Q,  and  the  plane  P  determined  by  AB  and  CD.    

To  prove  P  II  Q.  \^ V 

Proof.     Either  P  is  or  is  not  parallel  to  Q. 

If  P  is  not  parallel  to  Q,  it  will  intersect  it 
in  a  line  parallel  to  both  AB  and  CD.        §  555 

But  this  is  impossible.  §  118 

.-.  P  II  Q. 


286  SOLID  GEOMETRY 

566.  Theorem.  Through  a  given  point  one  plane,  and  only 
one  can  be  passed  parallel  to  any  given  plane  not  containing  the 
point. 

Pass  a  plane  through  the  given  point  and  intersecting  the  given  plane. 
Through  the  point  and  in  this  plane  draw  a  line  parallel  to  the  intersection 
of  the  two  planes.     This  line  is  parallel  to  the  given  plane  by"§  554. 

In  a  similar  manner  draw  a  second  line  through  the  given  point,  and 
apply  §  565. 

567.  Theorem.  If  two  angles,  not  in  the  same  plane,  have 
their  sides  parallel,  right  side  to  right  side  and  left  side  to  left  side, 
the  angles  are  equal  and  their  planes  are  parallel. 

Given  Zx  in  plane  P  and  Zz  in  plane  Q, 
having  AB  II  DE  and  CB  II  FE. 

To  prove  Zx  =  Zz,  and  P  II  Q. 

Proof.     Take    BA=ED,    and    BC  =  EF, 
and  draw  lines  AD,  CF,  BE,  AC,  and  DF. 

Both  AE  and  CE  are  parallelograms.  Why? 

Hence  both  AD  and  CF  are  equal  and  parallel  to  BE.  Why? 

Then  AD  and  CF  are  equal  and  parallel.  Why? 

Therefore  ADFC  is  a  parallelogram..  §  160 

And  AC  =  DF.  Why? 

Then  AABC^ADEF.  §  113 

.'.Zx  =  Zz.  Why? 

Also  AB  II  Q,  and  CB  II  Q.  §  554 

.'.  PWQ.  §  565 

EXERCISES 

1.  What  is  the  locus  of  a  line  that  passes  through  a  given  point,  and  i; 
parallel  to  a  given  plane? 

2.  If  two  congruent  parallelograms  lie  in  parallel  planes  and  hav 
their  sides  respectively  parallel,  how  many  planes  are  determined  in  usinj 
their  sides  in  all  possible  combinations? 

3.  The  four  lines  in  which  two  parallel  planes  intersect  two  other  par- 
allel planes  are  parallel. 

4.  If  two  planes  are  parallel  to  the  same  line,  their  intersections  with 
any  plane  through  the  line  are  parallel  to  each  other. 


PARALLEL  LINES  AND  PLANES 


287 


568.    Theorem.    //  two  straight  lines  are  cut  by  three  parallel 
planes,  their  corresponding  segments  are  proportional. 

-L 


\ 


.--7C\ 

/    M 


Wt 


Given  AB  and  CD  two  straight  lines  intersected  by  three 
parallel  planes  M,  P,  and  Q,  in  the  points  A,  E,  and  B,  and  C, 
F,  and  D  respectively. 
AE 


To  prove  —  = 


CF 
EB    FD 

Proof.     Draw  AD  intersecting  plane  P  in  G. 
Let  the  plane  determined  by  AB  and  AD  intersect  plane  P 
in  EG  and  plane  Q  in  BD. 

Then  EG  II  £D.  §  562 

Similarly,  if  AD  and  CD  determine  a  plane  intersecting  plane 

M  in  AC  and  plane  P  in  ffiF,  GF  II  AC.  Why? 

4^L4^  and—  =  — 
SB    GD' an    FZ)    GD 

.  4^  =  ^ 


Hence, 


§411 


Why? 


569.  Theorem.  If  two  straight  lines  are  cut  by  a  series  of 
parallel  planes,  the  corresponding  segments  are  proportional. 

Exercise.  A  mine  shaft  passes  for  180  ft.  on  a  slant  through  two  layers 
of  rock.  One,  a  layer  of  hard  rock,  is  72  ft.  thick;  and  the  other,  a  layer  of 
soft  rock,  is  84  ft.  thick.  If  it  costs  $4  per  linear  foot  in  the  hard  rock  and 
$3  in  the  soft  to  make  the  shaft,  find  the  cost  of  digging  the  shaft. 


288 


SOLID  GEOMETRY 


LINES  AND   PLANES,   PERPENDICULAR 

570.  Definition.     A  straight  line  is  perpendicular  to  a  plane 

if  perpendicular  to  all  straight  lines  in  the  plane  and  passing 

through  its  foot.     The  plane  is  also  perpendicular  to  the  line. 

A  straight  line  is  oblique  to  a  plane  when  it  meets  the  plane 

but  is  not  perpendicular  to  it. 

571.  Theorem.  //  a  line  is  perpendicular  to  each  of  two 
intersecting  lines  at  their  point  of  intersection,  it  is  perpendicular 
to  the  plane  determined  by  these  lines. 


-  ■ 

lf     :■* 

A 

? 

k 

» 

/ 

Given  AP  J_  AB  and  AD  at  their  point  of  intersection  A, 
and  plane  Q  determined  by  AB  and  AD. 

To  prove  AP  perpendicular  to  the  plane  Q. 

Proof.  In  plane  Q  draw  BD,  any  line  not  passing  through 
A,  and  through  A  draw  any  line  AC  meeting  BD  in  C. 

Prolong  PA  to  E,  making  AE  =  AP;  and  draw  BP,  CP,  DP, 
BE,  CE,  and  DE. 

Then  AB  and  AD  are  perpendicular  bisectors  of  PE. 


Therefore  BP  =  BE,  and  DP  =  DE. 

Then  ABPD^ABED. 

And  ZCBP  =  ZCBE. 

Further  ACBP^ACBE. 

And  therefore  CP  =  CE. 

Hence  AC  _L  PE,  i.e.,  AP  _L  AC,  any  line  in  plane  Q  and 
passing  through  its  foot.  Why? 

.\  AP  is  perpendicular  to  the  plane  Q.  §  570 


Why? 
§205 
§113 

Why? 
§247 

Why? 


PERPENDICULAR  LINES  AND  PLANES  289 

572.  Theorem.  All  the  perpendiculars  that  can  be  drawn  to 
a  straight  line  at  a  given  point  in  the  line,  lie  in  a  plane  perpen- 
dicular to  the  line  at  the  given  point. 

Given  DC  any  line  J_  AB  at  C,  and  plane    r— 


Q±ABatC.  \  e\ 1 

\    D 
To  prove  that  all  lines  perpendicular  to       \ — , r 

AB  at  C  lie  in  plane  Q.  ■ ' 


Proof.     DC  and  AB  determine  a  plane  P  that  intersects 
plane  Q  in  EC. 
Then  EC  JL  AB.  §  570 

But  DCA.AB.  Given 

Therefore  DC  coincides  with  EC.  Why? 

And  DC  lies  in  plane  Q. 

.*.  all  lines  J_  AB  at  C  lie  in  plane  Q. 

573.  Theorem.  Through  a  given  point  one  plane  can  be 
passed  perpendicular  to  a  given  straight  line,  and  only  one. 

If  the  given  point  is  in  the  given  line,  two  perpendiculars  to  the  line 
determine  a  plane  perpendicular  to  the  given  line.  If  the  given  point  is 
outside  the  given  line,  construct  a  perpendicular  from  the  point  to  the 
line,  and  at  its  foot  construct  another  perpendicular  to  the  given  line. 
These  determine  the  perpendicular  plane. 

To  show  that  only  one  plane  can  be  drawn  in  each  case,  suppose  that 
there  are  two  planes  perpendicular  and  pass  a  plane  through  the  given 
line  and  intersecting  the  two  planes  in  two  lines.     What  follows? 

EXERCISES 

1.  Show  how  a  perpendicular  to  a  plane  can  be  determined  by  means 
of  two  carpenter's  squares. 

2.  Each  of  three  concurrent  lines  is  perpendicular  to  each  of  the  other 
two.     Prove  that  each  is  perpendicular  to  the  plane  of  the  other  two. 

3.  If  a  book  partly  open  stands  on  a  table,  is  the  back  of  the  book 
perpendicular  to  the  table  top? 

4.  One  side  of  a  right  angle  revolved  about  the  other  side  as  an  axis 
generates  a  plane  perpendicular  to  this  other  side. 

5.  Do  the  hands  of  a  clock  revolve  in  a  plane?     Why? 


290 


SOLID  GEOMETRY 


574.  Theorem.  From  a  given  point  outside 
perpendicular  to  the  plane,  and  only  one,  can  be 

Given  plane  Q  and  point  A  outside  Q. 

To  prove  one  line  perpendicular  to 
Q  from  A,  and  only  one,  can  be  drawn. 

Proof.     In   plane   Q   draw  any  line 
CD.     Through  A  pass  a  plane  P  J_ 
CD.  §  573 

Let  BF  be  the  intersection  of  P  with  Q. 

Construct  AB  1  BF,  and  EF  II  AB. 

In  Q,  construct  GB  II  CF. 

Then  CF  _L  EF. 

And  ZABG  =  ZEFC. 

Hence  AB  J_  G£. 

Therefore  Ai?  J_  plane  Q. 

Only  one  perpendicular  can  be  drawn,  for 
drawn,  they  would  both  be  perpendicular  to 
of  their  plane  with  Q,  which  is  impossible. 


a  plane  one  line 
drawn. 


§§  312,  117 
§117 

Why? 
§567 

Why? 
§571 

if  two  could  be 

the  intersection 

Why? 


575.  Theorem.  At  a  point  in  a  plane  one  perpendicular  can 
be  drawn  to  the  plane,  and,  only  one. 

Let  0  be  the  point  in  plane  Q.     Draw  AB,  any 
line  in  Q  through  0.     Construct  plane  P  ±  AB       /c 
through  0.     In  P  construct  OE  _L  CD.     Complete 
and  give  proof. 

576.  Theorem.  The  perpendicular  to  a  plane  is  the  shortest 
line  from  a  point  to  a  plane. 

Apply  §  185. 

577.  Definition.  The  length  of  the  perpendicular  from  a 
point  to  a  plane  is  called  the  distance  from  the  point  to  the 
plane. 

i,    The  distance  between  two  parallel  planes  is  the  length  of 
the  segment  of  a  common  perpendicular  lying  between  them. 


PERPENDICULAR  LINES  AND  PLANES  291 

EXERCISES 

1.  What  is  the  locus  of  all  points  equidistant  from  a  circle? 

2.  What  is  the  locus  of  all  points  equidistant  from  the  vertices  of  a  tri- 
angle? 

3.  What  is  the  locus  of  all  points  equidistant  from  two  given  points? 

4.  Find  the  length  of  the  locus  in  a  plane  of  all  points  10  in.  from  a 
point  6  in.  from  the  plane. 

5.  With  a  pair  of  compasses  opened  13  in.  show  how  a  circle  of  12  in. 
radius  can  be  drawn. 

6.  A  perpendicular  smokestack  is  braced  by  three  guy  wires  attached 
at  a  point  90  ft.  from  the  ground.  If  these  wires  are  150  ft.  long  arid  meet 
the  ground  at  the  vertices  of  an  equilateral  triangle,  find  the  length  of  one 
side  of  the  triangle. 

7.  Could  a  smokestack  be  fully  braced  with  three  guy  wires?  With 
two?     Why? 

8.  If  from  the  foot  of  a  perpendicular  to  a  plane  a  line  is  drawn  perpen- 
dicular to  any  given  line  in  the  plane,  the  line  joining  the  point  of  inter- 
section to  any  point  in  the  perpendicular  to  the  plane  is  perpendicular  to 
the  given  line  in  the  plane. 

578.  Theorem.  Two  planes  perpendicular  to  the  same  straight 
line  are  parallel;  and  conversely,  if  one  of  two  parallel  planes  is 
perpendicular  to  a  straight  line,  the  other  is  also. 

Given  planes  P  and  Q  _L  AB.  I 

To  prove  P II  Q.  \P     c\         \ 

Outline   of  proof.     If  the   planes  were   not   

parallel,  they  would   meet.     Two   perpendicu-  \q     d\         \ 
lars  could  then  be  drawn  to  the  line  from   a  5 

point  in  their  intersection. 

Conversely: 

Given  plane  P  parallel  to  plane  Q,  and  P  ±  AB, 

To  prove  Q  _L  AB. 

Outline  of  proof.  AB  is  perpendicular  to  any  line  in  P 
through  its  foot.     Draw  two  such  lines. 

The  planes  determined  by  these  lines  and  AB  will  intersect 
Q  in  lines  parallel  respectively  to  the  lines  in  P,  and,  therefore, 
will  be  perpendicular  to  AB, 


292  SOLID  GEOMETRY 

579.  Theorem.  If  two  planes  are  parallel  to  a  third  plane, 
they  are  parallel  to  each  other. 

Proof  similar  to  that  of  §  124. 

580.  Theorem.  If  one  of  two  parallel  lines  is  perpendicular 
to  a  plane,  the  other  is  also;  and  conversely,  two  straight  lines  per- 
pendicular to  the  same  plane  are  parallel. 

Given  AB  II  CD,  and  AB  perpendicular 
to  plane  P. 

To  prove  CD  _L  P. 

Proof.     Through  D  draw  any  line  DF 
in  plane  P. 

Through  B  draw  BE  in  plane  P  and  parallel  to  DF. 

Then  ZA BE  =  ZCDF.  §  567 

But  ZABE  is  a  rt.Z.  Why? 

Therefore  ZCDF  is  a  rt.Z  and  CD  ±  DF,  any  line  in  plane  P 
through  D. 

.'.  CD  ±  P.  Why? 

Conversely:  a        c  e 

Given  AB  _L  plane  P,  and  CD  J_  plane  P. 

To  prove  AB  II  CD. 

Outline  of  proof.  Through  D  draw  ED  II  . 
AB.  Then  ED  ±  P,  and  therefore  coincides/ 
with  CD  by  §  575. 


D 

P/ 


581.  Theorem.  Two  parallel  planes  are  everywhere  the  same 
distance  apart. 

Choose  any  two  points  in  one  plane  and  draw  perpendiculars  to  the  other 
plane,  §  574.  Then  these  lines  are  perpendicular  to  the  first  plane,  §  578. 
Further  these  two  lines  are  parallel,  §  580,  and  determine  a  plane 
intersecting  the  two  parallel  planes  in  parallel  lines,  §  562.  Then  apply 
§  191. 

EXERCISES 

1.  Prove  the  theorem  of  §  553  by  §  580. 

2.  A  line  cannot  be  perpendicular  to  each  of  two  intersecting  planes. 


PROJECTIONS 


293 


A 

p 

/Uj 

\ 

i 

B 

X 

Y^> 

\ 

PROJECTIONS 

582.  Definitions.  The  projection  of  a  point  upon  a  plane 
is  the  foot  of  the  perpendicular  from  the  point  to  the  plane. 
The  projection  of  a  line  upon  a  plane  c 
is  the  locus  of  the  projections  of  all  points 
of  the  line  upon  the  plane. 

Thus  B  is  the  projection  of  point  A  upon  plane 
Q,  and  EF  is  the  projection  of  line  CD. 

583.  Theorem.  The  projection  upon  a  plane  of  a  straight  line 
that  is  not  perpendicular  to  the  plane  is  a  straight  line. 

Given  AB,  a  straight  line  not  _L  plane  Q. 

To  prove  that  the  projection  of  AB  upon 
Q  is  a  straight  line. 

Proof.     From  any  point  in  AB,  as  E, 
draw  EF  ±Q.  §  574 

Plane  P,  determined  by  AB  and  EF,  intersects  Q  in  CD.  §  551 

From  G,  any  other  point  in  AB,  draw  GH  J_  Q.  §  574 

Then  GH  II  EF  and  lies  in  P.  Why? 

Therefore  its  foot  H  lies  in  CD.  Why? 

And  the  projections  of  all  points  of  AB  lie  in  CD. 

Furthermore,  a  perpendicular  to  Q  at  any  point  in  CD  will 
intersect  AB.  Why? 

Then  every  point  in  CD  is  the  projection  of  a  point  in  AB. 
.'.  the  projection  of  AB  upon  Q  is  a  straight  line. 

584.  Theorem.  Of  all  oblique  lines  drawn  from  a  point  to  a 
plane: 

(1)  Those  that  have  equal  projections  are  equal. 

(2)  Those  that  have  unequal  projections  are  unequal,  and  the 
one  having  the  greater  projection  is  the  greater. 

Another  statement  of  this  theorem  is:  Of  all  oblique  lines,  drawn  to  a 
plane  from  a  point  in  a  perpendicular  to  the  plane,  those  that  cut  off 
equal  distances  from  the  foot  of  the  perpendicular  are  equal,  and  of  those 
that  cut  off  unequal  distances  from  the  foot  of  the  perpendicular,  the 
more  remote  is  the  greater. 


294  SOLID  GEOMETRY 

585.  Theorem.  The.  acute  angle  that  a  straight  line  makes 
with  its  projection  upon  a  plane  is  the  least  angle  that  it  makes 
with  any  line  passing  through  its  foot  and  lying  in  the  plane. 

Given  the  straight  line  AB  meeting  plane 
Q  at  B,  its  projection  CB  in  Q,  and  DB,  any- 
other  line  through  B  and  in  Q. 

To  prove  ZABC  <  ZABD. 

Proof.  Make  BD  =  BC,  and  draw  AC 
and  AD. 

Then  ZABC  <  ZABD.  §  259 

586.  Definition.  The  acute  angle  that  a  straight  line  makes 
with  its  own  projection  upon  a  plane  is  called  the  inclination 
of  the  line  to  the  plane,  or  the  angle  that  the  line  makes 
with  the  plane. 

EXERCISES 

1.  State  and  prove  the  converse  theorems  to  §  584. 

2.  How  does  the  length  of  the  projection  of  a  line  upon  a  plane  com- 
pare with  the  length  of  the  line  when:  (1)  the  line  is  parallel  to  the  plane; 
(2)  the  line  is  perpendicular  to  the  plane;  (3)  the  line  is  neither  parallel 
nor  perpendicular  to  the  plane? 

3.  Which  is  the  greatest  angle  that  an  oblique  line  to  a  plane  makes 
with  any  line  in  the  plane  and  through  its  foot? 

4.  Parallel  lines  make  equal  angles  with  a  plane. 

5.  If  a  straight  line  intersects  two  parallel  planes  it  makes  equal  angles 
with  the  parallel  planes. 

6.  Find  the  projection  of  a  line  16  in.  long  upon  a  plane  if  the  angle  it 
makes  with  the  plane  is  45°.     If  30°.     If  60°. 

7.  What  is  the  locus  of  a  point  in  a  given  plane  and  equidistant  from 
two  given  points  not  in  the  given  plane? 

8.  A  rectangle  8  in.  by  12  in.  is  intersected  along  one  of  its  diagonals 
by  a  plane.  If  the  other  diagonal  makes  an  angle  of  45°  with  the  plane, 
find  the  distance  from  the  extremities  of  this  diagonal  to  the  plane. 

Ans.     5.098  in. 

9.  The  projection  of  a  given  line  on  each  of  two  planes  is  a  in  length. 
Are  the  two  planes  necessarily  parallel?     Illustrate. 


DIEDRAL  ANGLES 


295 


DIEDRAL  ANGLES 

587.  Definitions.  An  angle  formed  by  two  intersecting 
planes  is  called  a  diedral  angle.  The  planes  are  called  the 
faces  of  the  diedral  angle,  and  their  intersection  is  called  the 
edge. 

If  a  plane  Q  revolves  about  an  axis  AB, 
and  if  a  line  CD  _L  AB  is  taken  in  Q,  then 
the  plane  Q  generates  a  diedral  angle,  and 
the  line  CD  generates  a  plane  angle  since 
it  revolves  in  a  plane.  When  the  plane 
angle  is  acute,  right,  obtuse,  etc.,  the 
diedral  angle  generated  in  connection 
with  it  is  acute,  right,  obtuse,  etc.  That  is,  the  diedral  angle 
and  the  plane  angle  are  each  formed  by  the  same  amount  of 
turning. 

588.  Plane  Angle.  The  plane  angle  generated  by  CD  is 
called  the  plane  angle  of  the  diedral  angle.  It  is  formed  by 
two  lines,  one  in  each  face  of  the  diedral  angle,  and  perpen- 
dicular to  the  edge  at  the  same  point. 

A  diedral  angle  is  read  by  naming  its  faces,  as  the  diedral  angle  QP;  or 
it  may  be  read  C-AB-D;  or  simply  AB. 

The  words  complementary,  supplementary, 
adjacent,  vertical,  etc.,  when  applied  to  diedral 
angles,  have  meanings  corresponding  to  their 
meanings  when  applied  to  plane  angles. 

589.  The  following  facts  are  readily 
deduced  from  the  definitions:  a 

(1)  All  plane  angles  of  a  diedral  angle  are  equal. 

(2)  If  two  diedral  angles  are  equal,  their  plane  angles  are  equal; 
and  conversely. 

(3)  Two  diedral  angles  are  proportional  to  their  plane  angles; 
that  is,  the  plane  angle  of  a  diedral  angle  can  be  taken  as  the  measure 
of  the  diedral  angle. 


296 


SOLID  GEOMETRY 


EXERCISES 

1.  By  comparison  with  the  corresponding  terms  in  plane  geometry, 
form  definitions  applying  to  diedral  angles  for  right,  oblique,  vertical, 
adjacent,  supplementary,  alternate  interior.  How  many  of  these  can  be 
illustrated  by  the  leaves  of  an  open  book? 

2.  With  reference  to  planes,  what  corresponds  to  the  following  axiom 
in  plane  geometry:  Through  a  given  point  .only  one  straight  fine  can  be 
drawn  parallel  to  another  straight  line? 

3.  If  one  plane  intersects  another  plane,  the  vertical  diedral  angles  are 
equal. 

4.  If  the  sum  of  two  adjacent  diedral  angles  is  two  right  diedral  angles, 
the  exterior  faces  lie  in  the  same  plane. 

5.  If  two  planes  are  cut  by  a  third  plane  so  as  to  make  the  alternate 
diedral  angles  equal,   the  two  planes  are 

parallel. 

6.  Take  a  piece  of  heavy  paper,  rule 
it  as  shown  in  figure  (1),  and  cut  along 
AO.  Fold  as  shown  in  (2)  and  show  how 
it  may  be  used  in  measuring  diedral  angles. 


w. 


U) 


(2) 


590.  Perpendicular  Planes.  If  two  planes  form  a  right 
diedral  angle,  the  two  planes  are  said  to  be  perpendicular  to 
each  other. 


591.    Theorem.     If  a  line  is  perpendicular  to  a  plane,  every 
plane  containing  this  line  is  perpendicular  to  the  given  plane. 

Given  CD  perpendicular  to  plane  PQ,  and  plane  MN  through 
CD  intersecting  PQ  in  AB. 

To  prove  that  plane  MN  is  perpendic- 
ular to  plane  PQ. 

Proof.     CD  _L  AB.  Why? 

In  plane  PQ  through  D,  draw  DELAB. 

Then  ZCDE  is  the  plane  angle  of  the  -P^ 
diedral  ZM-AB-Q.  9  §  588 

But  ZCDE  is  a  rt.Z.  Why? 

Hence  diedral  ZM-AB-Q  is  a  rt.  diedral  Z.  §  589  (3) 

.*.  plane  MN  is  perpendicular  to  plane  PQ.        Why? 


DIEDRAL  ANGLES  297 

592.    Theorem.     //  a  straight  line  is  parallel  to  a  plane,  any 
plane  perpendicular  to  the  line  is  perpendicular  to  the  plane. 


Given  AB  II  plane  M,  and  plane  P  _L  AB  at  C. 
To  prove  P  _L  M . 

Proof.     From  any  point  B  in  AB  draw  BG  _L  M.  §  574 

AB  and  BG  determine  a  plane  Q  which  intersects  plane  M 
in  DG,  and  plane  P  in  CD. 

Further                 DG  II  AB,  and  CD  _L  CB.  Why? 

Then                                   CD   II  BG.  Why? 

Therefore                           CD  1  M.  §  580 

.-.  P1M.  §  591 

EXERCISES 

1.  Through  a  given  point  pass  a  plane  perpendicular  to  a  given  plane. 
How  many  such  planes  can  be  drawn? 

2.  If  one  line  is  perpendicular  to  another  line,  is  every  plane  containing 
the  first  line  perpendicular  to  the  second?     Prove. 

3.  A  plane  determined  by  a  line  and  its  projection  upon  a  given  plane 
is  perpendicular  to  the  given  plane. 

4.  If  three  concurrent  lines  are  each  perpendicular  to  the  other  two, 
what  can  be  proved  concerning  the  planes  determined  by  the  lines?  Illus- 
trate by  the  walls  of  a  room. 

5.  If  a  plane  is  perpendicular  to  a  line  in  another  plane  it  is  perpen- 
dicular to  that  other  plane. 

6.  State  the  converse  of  the  theorem  of  §  592.  Is  it  a  true  theorem? 
Prove. 

7.  A  circle  is  divided  into  eight  equal  parts  by  diameters.  A  plane 
containing  one  of  these  diameters  makes  an  angle  of  45°  with  the  plane  of 
the  circle.  Find  the  distance  of  each  division  point  of  the  circle  from  the 
plane. 


298  SOLID  GEOMETRY 

593.  Theorem.  If  two  planes  are  perpendicular  to  each  other 
a  line  drawn  in  one  of  them  perpendicular  to  their  intersection  is 
perpendicular  to  the  other. 

Given  plane  MN  J_  plane  PQ,  AB  their  intersection,  and 
CD  ±  AB  in  plane  MN. 

To  prove  CD  _L  plane  PQ. 

Proof.     Through  D  and  in  plane  PQ  draw  DE  _L  AB. 

Then  ZCDE  is  the  plane  angle  of  the  right  diedral  ZM-AB-Q. 


Hence  ZCDE  is  a  rt.Z,  and  CD  _L  DE.  Why? 

But  CD  J_  AB.  Given 

.\  CD  _L  plane  PQ.  §  571 

594.  Theorem.  If  two  planes  are  perpendicular  to  each  other 
a  perpendicular  to  one  of  them  at  any  point  of  their  intersection  will 
lie  in  the  other. 

Given  plane  MN  _L  plane  PQ,  and  intersecting  it  in  AB, 
afso  CD  JL  plane  PQ  at  any  point  D  in  AB. 

To  prove  that  CD  must  lie  in  plane  MN. 

Suggestion.  In  plane  MN  draw  a  line  _L  AB  at  D.  Then 
this  line  is  J_  plane  PQ,  and  must  coincide  with  CD.  Why? 

595.  Theorem.  If  two  planes  are  perpendicular  to  each  other 
a  perpendicular  to  one  from  any  point  in  the  other  will  lie  in  the 
other. 

Given  plane  MN  J_  plane  PQ,  and  intersecting  it  in  AB, 
also  CD  J_  plane  PQ  from  any  point  C  in  plane  MN. 

To  prove  that  CD  must  lie  in  plane  MN. 

Suggestion.  In  plane  MN  draw  a  line  _L  AB  from  C.  Then 
this  line  is  _L  plane  PQ,  and  must  coincide  with  CD. 


DIEDRAL  ANGLES 


299 


596.  Theorem.  If  each  of  two  intersecting  planes  is  perpen- 
dicular to  a  third  plane  their  line  of  intersection  is  perpendicular 
to  the  third  plane.  ^    H 

Given  planes  P  and  Q,  intersecting  ^Til^"H^ 

in  AB,  and  each  _L  plane  M. 
To  prove  AB  ±  M. 
Proof.     Either  AB  _L  M  or  it  is  not. 
If  AB  is  not  J_  M,  draw  a  line  HB  ±  M  at  B. 
Then  HB  lies  in  both  P  and  Q.  §  594 

Hence  HB  is  the  intersection  of  P  and  Q.  Why? 

But  Ai?  is  the  intersection  of  P  and  Q.  Given 

Therefore  A B  and  i7J5  coincide.  §  551 

.-.  AB  J_  AT. 

597.  Theorem.  Through  any  straight  line  not  perpendicular 
to  a  plane,  one  plane,  and  only  one,  can  be  passed  perpendicular 
to  the  given  plane. 

Given  plane  P  and  AB  a  line  not  J_  P. 

To  prove   that  one  plane  _L  P  can  be 
passed  through  AB,  and  only  one. 

Proof.     From  any  point  A  in  AB,  draw  AC  LP. 

AB  and  AC  determine  a  plane  JL  P.  §  591 

.*.  one  plane  _L  P  can  be  passed  through  AB. 

If  another  plane  _1_  P  could  be  drawn  through  AB,  then  AB 
would  be  ±  P  by  §  596. 

But  AB  is  not  J_  P.  Given 

.*.  only  one  plane  J_  P  can  be  passed  through  AB. 


EXERCISES 

1.  Discuss  the  theorem  of  §  597  for  the  case  where  AB  lies  in  plane  P. 

2.  If  a  plane  is  perpendicular  to  each  of  two  intersecting  planes,  it  is 
perpendicular  to  their  intersection. 

3.  Any  point  in  a  plane  containing  the  bisector  of  an  angle,  and  per- 
pendicular to  the  plane  of  the  angle,  is  equally  distant  from  the  sides  of 
the  angle. 


300 


SOLID  GEOMETRY 


598.    Theorem.     Between  two  straight  lines  not  in  the  same 
plane  one  common  perpendicular,  and  only  one,  can  be  drawn. 


Given  AB  and  CD  two  non-coplanar  straight  lines. 

To  prove  that  one  common  perpendicular  to  AB  and  CD,  and 
only  one,  can  be  drawn. 

Proof.     Through  AB  pass  plane  P  II  CD.  §  559 

Through  CD  pass  plane  EH  JL  P,  and  intersecting  P  in  GH. 

§597 

Then  GH  II  CD.  Why? 

And  GH  intersects  AB  in  some  point  0.  Why? 

In  plane  EH  draw  NO  _L  GH  at  0. 

Then  NO  _L  CD  and  M)  _L  AB.  Why? 

.*.  one  common  perpendicular  to  AB  and  CZ)  can  be  drawn. 

Suppose  that  there  is  another  common  perpendicular  KL. 

In  plane  P  draw  LS  II  GH,  and  in  plane  EH  draw  2J72  JL  GH. 

Then  LS  II  CD,  and  hence  KL  ±  LS.  Why? 

Therefore  KL  J_  plane  P.  Why? 

But  this  is  impossible  for  KR  J_  plane  P.  Why? 

Hence  KL  is  not  _L  AB. 

,\  only  one  common  perpendicular  to  AB  and  CD  can  be 
drawn. 

599.  Theorem.  The  common  perpendicular  between  two  non- 
coplanar  lines  is  the  shortest  line  that  can  be  drawn  between  them. 

Prove  that  NO  is  less  than  any  other  line  KL,  for  NO  =  KR  and  KR  <  KL . 

600.  Definition.  The  length  of  the  shortest  line  between 
two  lines  is  called  the  distance  between  them. 

Compare  with  §§  187-190,  577. 


DIEDRAL  ANGLES 


301 


601.    Theorem.    Every  point  in  a  plane  that  bisects  a  diedral 
angle  is  equidistant  from  the  faces  of  the  angle. 


Given  the  plane  R  bisecting  the  diedral  angle  formed  by  the 
planes  P  and  Q,  and  N  any  point  in  plane  R.  Also  NC  _L  P 
and  ND  ±  Q. 

To  prove  NC  =  ND. 

Proof.     NC  and  ND  determine  plane  CD.  Why? 

Plane  CD  intersects  P  in  CO,  Q  in  DO,  and  R  in  NO. 

Plane  CD  LP  and  also  1  Q.  §  591 

Then  plane  CD  ±  AB.  Why? 

And  NO,  CO,  and  DO  are  each  _L  AB.  Why? 

Therefore  ZNOC  is  the  plane  angle  of  diedral  ZRP,  and 
/.NOD  is  the  plane  angle  of  diedral  ZRQ. 

Hence  ZNOC  =  ZNOD.  Why? 

Show  ANOC^ANOD. 

.'.NC  =  ND.  Why? 

602.  Theorem.  Every  point  equidistant  from  the  two  faces  of 
a  diedral  angle  lies  in  the  plane  bisecting  the  angle. 

EXERCISES 

1.  Are  both  of  the  preceding  theorems  included  in  the  following?  The 
locus  of  a  point  equidistant  from  the  faces  of  a  diedral  angle  is  the  plane 
bisecting  the  angle.     Explain. 

2.  From  any  point  within  a  diedral  angle  perpendiculars  are  drawn 
to  the  faces.  Prove  that  the  angle  formed  by  these  perpendiculars  is 
supplementary  to  the  plane  angle  of  the  diedral  angle. 

3.  The  plane  angle  of  a  diedral  angle  is  120°.  A  point  in  the  bisector  of 
the  diedral  angle  is  16  in.  from  the  edge  of  the  angle.  Find  the  distance  of 
this  point  from  the  faces  of  the  angle. 


302 


SOLID  GEOMETRY 


POLYEDRAL  ANGLES 


603.  Definitions.     When  three  or  more  planes  meet  at 
point  they  form  a  polyedral  angle,  or  a  solid  angle. 

The  polyedral  angle  is  formed  by  a  portion 
of  the  planes  as  shown  in  the  figure.  The  point 
V  in  which  all  the  planes  meet  is  the  vertex  of 
the  polyedral  angle;  the  lines,  AV,  BV,  and  CV, 
of  intersection  of  consecutive  planes  are  the 
edges;  the  portions  of  the  planes  lying  between 
the  edges  are  the  faces;  and  the  angles  in  the  faces  between 
the  edges  are  the  face  angles. 

A  polyedral  angle  is  read  by  the  letter  at  the  vertex,  or  by 
this  letter  together  with  a  letter  on  each  edge. 

Thus,  the  polyedral  angle  in  the  figure  is  read  "polyedral 
angle  7,  or  V-ABC" 

604.  Parts  of  a  Polyedral  Angle.  The  face  angles  are 
AVB,  BVC,  and  CVA.  The  diedral  angles  formed  by  the 
faces  are  the  diedral  angles  of  the  polyedral  angle.  The  parts 
of  a  polyedral  angle  are  its  face  angles  and  its  diedral  angles. 

For  convenience  in  representing  a  polyedral  angle,  a  plane 
is  often  passed  through  it  cutting  all  its  edges.  It  should  be 
noted  that  this  plane  is  not  a  part  of  the  polyedral  angle 

605.  Convex  and  Concave  Polyedral  Angles.  If  a  plane 
cuts  all  the  edges  of  a  polyedral  angle,  but  not  through  the 
vertex,  the  intersections  of  the 
faces  with  this  plane  form  a 
polygon.  If  this  polygon  is  con- 
vex (§  169),  the  polyedral  angle 
is  said  to  be  convex.  If  the 
polygon  is  concave,  the  poly- 


CONVEX 


COJSCAVE 


edral  angle  is  said  to  be  concave. 

In  this  text  only  convex  polyedral  angles  will  be  considered, 
unless  otherwise  stated. 


POLYEDRAL  ANGLES 


303 


606.  Classification.  A  polyedral  angle  that  has  three  faces 
is  called  a  triedral  angle.  The  words  tetraedral,  pentaedral, 
hexaedral,  etc.,  may  be  applied  when  the  polyedral  angle  has 
four,  five,  six,  etc.,  faces. 

607.  Congruent  Polyedral  Angles.  If  the  corresponding 
parts  of  two  polyedral  angles  are  equal  and  arranged  in  the 
same  order,  the  polyedral  angles  are  congruent. 

Thus,  in  the  figure,  the  face  angles 
are  equal  and  arranged  in  the  same 
order,  that  is,  ZAVB  =  ZA'V'B', 
ZBVC  =  ZB'V'C,  and  ZCVA  = 
ZC'Y'A1.  Also  in  the  diedral  angles, 
ZAV  =  ZA'V,  ZBV  =  ZB'V,  and 
ZCV  =  ZCfV,  the  arrangement  is  in  the  same  order. 

If  the  corresponding  parts  of  two  polyedral  angles  are  equal 
and  arranged  in  the  opposite  order,  the  polyedral  angles  are 
said  to  be  symmetric.     These  will  be  considered  later  (§  836). 


EXERCISES 

1.  Does  the  size  of  a  polyedral  angle  depend  upon  the  lengths  of  its 
edges?  How  do  the  number  of  edges,  the  number  of  diedral  angles,  and 
the  number  of  face  angles  of  a  polyedral  angle  compare? 

2.  Construct  from  stiff  paper  a  triedral  angle  having  face  angles  equal 
to  50°,  70°,  and  90°.  Can  you  tell  the  number  of  degrees  in  the  diedral 
angles? 

The  paper  may  be  cut  as  indicated  in  the 
figure,  folded  along  the  dotted  lines  and  pasted. 

3.  How  many  degrees  in  each  of  the  face 
angles  and  the  diedral  angles  of  the  polyedral 
angle  formed  by  the  walls  of  a  room? 

4.  The  face  angles  and  diedral  angles  of  a 
convex  polyedral  angle  are  each  less  than  180°. 

5.  Bearing  the  plane  geometry  definitions  in  mind,  define  vertical 
polyedral  angles.  Are  they  congruent  or  symmetric?  Explain  how,  if 
it  is  possible  to  have  two  vertical  triedral  angles  congruent. 

6.  Could  a  triedral  angle  have  one  right  diedral  angle?  Could  it  have 
two?     Three?     Give  illustrations. 


304  SOLID  GEOMETRY 

608.    Theorem.     The  sum  of  two  face  angles  of  a  triedra 
angle  is  greater  than  the  third  face  angle. 

Given  the  triedral  ZV-ABC. 

To  prove  that  the  sum  of  two  face  angles  is  greater  tha 

the  third  face  angle.  y 

Proof.     Three   cases   arise:    (1)  all  the  face         //i\ 

angles  equal;  (2)  two  face  angles  equal;  (3)  no  \  \    \ 

two  face  angles  equal.  A/  \\d 

Cases  (1)  and  (2)  are  left  to  the  student.  j^LZ-^^ 

In  case  (3)  suppose  ZAVB  is  the  greatest.  / 

In  the  face  A VB,  construct  ZAVD  =  ZAVC,  take  VD  =  VC 

and  draw  AB,  AC,  and  BC. 
Then  AA  VC  £ AA  VD.  §  24 

And  AC  =  AD.  Why** 

AC+CB>AD+DB.  §  182J 

Hence  CB>DB.  §  174J 

In  ACVB  and  DVB,  BV  =  BV,  and  CV  =  DV,  but  CB>DB. 
Therefore  ZCVB  >  ZDVB.  §  259 

Then  ZAVC + ZCVB  >ZAVD+ ZDVB.  §  173 

.-.  ZAVC+ZCVB>ZAVB. 
Or  the  sum  of  two  face  angles  is  greater  than  the  third. 


• 


EXERCISES 

1.  Any  face  angle  of  a  triedral  angle  is  greater  than  the  difference 
the  other  two. 

2.  State  the  theorems  in  plane  geometry  that  correspond  to  the  theorems 
of  §  608  and  Exercise  1. 

There  is  a  close  analogy  between  the  plane  triangle  and  the  triedrj 
angle.     Many  theorems  concerning  plane  triangles  can  be  changed  t 
theorems  concerning  triedral  angles  by  replacing  the   word   side  by  face 
angle,  and  the  word  angle  by  diedral  angle. 

3.  State  theorems  for  triedral  angles  analogous  to  the  theorems  of  plan 
geometry  that  have  to  do  with  congruent  triangles. 

4.  Referring  to  the  figure   of   §608,  ZAVC  =  55°   and   ZCVB  =  65 
Make  a  statement  regarding  the  number  of  degrees  in  ZAVB. 


POLYEDRAL  ANGLES  305 

609.    Theorem.     The  sum  of  the  face  angles  of  any  convex 
polyedral  angle  is  less  than  four  right  angles. 


Given  the  convex  polyedral  ZV-ABCDEF. 

To  prove  ZAVB+ZBVC+  •  ■  - ZFVA  <4  rt.A. 

Proof.  Let  a  plane  intersect  the  edges  of  the  polyedral 
angle  in  A,  B,  C,  etc.,  and  intersect  the  faces  in  AB,  BC,  CD, 
etc.  Join  any  point  in  the  polygon  thus  formed  to  the  vertices 
of  the  polygon. 

Then  ZVBA  +  ZVBOZABC.  §  608 

Similarly  ZVCB+ZVCD>ZBCD,  etc. 

Add  these  and  the  result  is  that  the  sum  of  all  the  base  angles 
of  the  triangles  with  vertices  at  V  is  greater  than  the  sum  of  all 
the  base  angles  of  the  triangles  with  vertices  at  0. 

Then,  since  the  sum  of  all  the  angles  of  the  triangles  with 
vertices  at  V  is  equal  to  the  sum  of  those  with  vertices  at  0, 
the  sum  of  the  angles  about  V  is  less  than  the  sum  of  the  angles 
about  0.  §  178 

But  the  sum  of  the  angles  about  0  =  4  xtA.  Why? 

.-.  ZAVB+ZBVC+  ■  •  •  ZFVA  <4  rt.A 

EXERCISES 

1.  Can  a  polyedral  angle  have  for  its  face  angles  three  angles  from  an 
equilateral  triangle?     Can  it  have  four?     Five?     Six?     Why? 

2.  Can  a  polyedral  angle  have  as  face  angles  the  angles  of  a  square? 
Of  a  regular  pentagon?     Of  a  regular  hexagon?     Why? 

3.  Any  face  angle  of  a  polyedral  angle  is  less  than  the  sum  of  the  others. 

4.  The  sum  of  the  face  angles  of  a  polyedral  angle  is  300°.  What  is 
the  greatest  value  that  any  one  of  the  face  angles  may  have? 

5.  One  of  the  face  angles  of  a  triedral  angle  is  60°.  What  are  the 
limitations  on  the  size  of  each  of  the  other  face  angles? 


306 


SOLID  GEOMETRY 


610.  Theorem.  Two  triedral  angles  are  congruent  if  the  three 
face  angles  of  one  are  equal  respectively  to  the  three  face  angles  of  the 
other,  and  arranged  in  the  same  order. 

Given  triedral  angles  V 
and  V,  having  ZAVB  = 
ZA'V'B',  ZBVC  =  ZB'V'C, 
ZCVA  =  ZC'V'A',  and  ar- 
ranged in  the  same  order. 

To  prove  triedral  ZV  =  triedral  ZV. 

Proof.  On  the  edges  of  triedral  angles  V  and  V  lay  off  the 
six  equal  segments  VA,  VB,  VC,  V'A',  V'B',  and  VC,  and 
draw  AB,  BC,  CA,  A'B',  B'C,  and  C'A'. 

Then  there  are  three  pairs  of  congruent  isosceles  triangles  as 
follows: 


Why? 
Why? 
Why? 
Why? 


AAVB^AA'VB',  ABVC^AB'VC, 
ACVA^AC'VA'. 

Hence       AB=A'B',  BC=B'C,  and  CA  =  C'A'. 

And  AABCtSAA'B'C. 

Also  A  VAC  and  VAB  are  acute. 

From  any  point  D  in  VA  construct  DE  in  face  AVB  and  DF 
in  face  AVC,  each  J_  VA.  These  lines  meet  AB  and  AC 
respectively  in  E  and  F.  Why? 

Draw  EF. 

Take  V'D'=VD  and  in  a  similar  manner  construct,  D'E', 
D'F',  and  E'F'. 

Then,   since  AD=A'D'  and   ZDAE=ZD'A'E', 
AADE^AA'D'E' 

Hence  AE =A'E'  and  DE = D'E'. 

Similarly  AF=A'F'  and  DF = D'F'. 

Therefore  AAEF^AA'E'F'. 

And  EF=E'F'. 

Therefore  ADEF^AD'E'F'. 

And  ZEDF  =  ZE'D'F'. 

Hence  diedral  ZVA  =  diedral  ZV'A'. 

Similarly,  prove  the  other  two  pairs  of  diedral  angles  equal. 
.-.  triedral  ZFstriedral  ZV.  §  607 


§97 
Why? 

§247 
Why? 

§113 

Why? 

§  589  (3) 


LOCI  307 

LOCI 

611.  As  has  been  assumed  on  the  previous  pages,  the  idea  of 
a  locus  here  is  the  same  as  in  plane  geometry. 

Here,  too,  the  proof  of  a  locus  theorem  must,  as  in  plane 
geometry,  consist  of  two  parts: 

(1)  That  all  points  in  the  figure  satisfy  the  given  conditions. 

(2)  That  all  points  that  satisfy  the  given  conditions  are  in  the 
figure. 

In  the  place  of  (2)  one  may  prove  that  all  points  not  in  the 
figure  do  not  satisfy  the  given  conditions. 

612.  In  plane  geometry  it  was  noticed  that  usually  a  locus 
satisfying  one  condition  is  a  line  or  group  of  lines,  and  a  locus 
satisfying  two  conditions  is  a  point  or  group  of  points.  There, 
however,  a  further  condition  is  assumed,  namely,  that  the  locus 
is  confined  to  a  plane. 

In  solid  geometry  usually,  one  condition  will  confine  the  locus 
to  a  surface  or  group  of  surfaces,  two  conditions  will  confine  the 
locus  to  a  line  or  group  of  lines,  and  three  conditions  will  confine 
the  locus  to  a  point  or  group  of  points. 

Here,  as  in  plane  geometry,  one  of  the  greatest  benefits 
derived  from  the  study  of  loci  is  through  the  imaging  and  con- 
structing figures,  rather  than  through  the  proofs  of  the  loci 
theorems.  This,  however,  does  not  mean  that  the  proofs 
should  be  neglected. 

EXERCISES 

1.  What  is  the  locus  of  points  equally  distant  from  two  given  points? 

2.  What  is  the  locus  of  points  equally  distant  from  three  given  points? 

3.  What  is  the  locus  of  points  equally  distant  from  four  given  points? 

4.  What  is  the  locus  of  points  equally  distant  from  two  intersecting 
lines  and  in  a  given  plane? 

5.  What  is  the  locus  of  the  end  point  of  a  line-segment  of  fixed  length, 
that  moves  so  as  to  remain  parallel  to  a  given  line  and  have  one  end  in  a 
given  plane? 


308  SOLID  GEOMETRY 

6.  What  is  the  locus  of  points  in  a  given  plane  equally  distant  from 
three  given  points  not  in  a  line? 

7.  What  is  the  locus  of  points  equally  distant  from  two  given  parallel 
lines? 

8.  What  is  the  locus  of  the  middle  point  of  a  line-segment  of  fixed 
length,  that  moves  so  as  to  have  its  end  points  in  two  parallel  planes? 

9.  What  is  the  locus  of  a  point  equally  distant  from  two  parallel 
planes  and  equally  distant  from  the  faces  of  a  diedral  angle? 

10.  What  is  the  locus  of  all  lines  that  pass  through  a  given  point  and 
are  parallel  to  a  given  plane  not  containing  the  point? 

11.  What  is  the  locus  of  the  points  within  a  diedral  angle  and  equally 
distant  from  its  two  faces? 

12.  What  is  the  locus  of  all  points  that  are  twice  as  far  from  a  line  in  a 
plane  as  from  the  plane? 

13.  What  is  the  locus  of  the  points  within  a  triedral  angle  and  equally 
distant  from  its  three  faces? 

14.  What  is  the  locus  of  the  points  equally  distant  from  the  edges  of  a 
triedral  angle? 

15.  What  is  the  locus  of  a  line  making  equal  angles  with  each  of  two 
intersecting  lines?  What  other  exercise  in  this  list  has  the  same  locus  as 
this? 

16.  What  is  the  locus  of  points  in  one  of  two  given  non-coplanar  lines 
equally  distant  from  two  given  points  in  the  other? 

17.  What  is  the  locus  of  a  point  that  moves  so  that  its  distance  from 
one  of  two  given  parallel  planes  is  to  its  distance  from  the  other  as  1  :  3? 

18.  What  is  the  locus  of  a  point  whose  distances  from  the  faces  of  a 
diedral  angle  are  respectively  5  in.  and  8  in.? 

19.  What  is  the  locus  of  the  projections  of  a  given  point  upon  the 
planes  containing  a  given  line? 

Many  questions  concerning  loci  cannot  be  discussed  in  elementary 
geometry  as  they  lead  to  forms  not  there  considered,  for  in  elementary 
geometry  only  a  limited  number  of  forms  such  as  lines,  circles,  planes, 
cones,  cylinders,  and  spheres  are  considered.  The  question:  What  is  the 
locus  of  the  middle  points  of  all  transversals  of  two  non-coplanar  lines? 
leads  to  none  of  these. 

In  general,  the  locus  of  points  equally  distant  from  two  different  ele- 
ments, as  a  point  and  a  line,  a  line  and  a  plane,  or  two  non-coplanar  lines 
cannot  be  considered  in  elementary  geometry. 


QUESTIONS  309 

QUESTIONS 

1.  State  the  different  relations  that  two  lines  may  have  to  each  other. 
That  a  line  may  have  to  a  plane.  That  two  planes  may  have  to  each 
other. 

2.  State  the  different  ways  of  determining  that  a  straight  line  lies  in 
a  plane. 

3.  What  is  the  intersection  of  a  line  and  a  plane?  Of  two  planes? 
Of  three  planes? 

4.  State  the  different  ways  of  determining  that  a  line  is  parallel  to  a 
plane. 

5.  State  the  different  ways  of  determining  that  a  line  is  perpendicular 
to  a  plane. 

6.  State  the  different  ways  of  determining  that  two  planes  are  parallel. 

7.  State  the  different  ways  of  determining  that  two  planes  are  perpen- 
dicular. 

8.  How  many  intersecting  lines  can  be  parallel  to  the  same  line?  To 
the  same  plane? 

9.  How  many  intersecting  planes  can  be  parallel  to  the  same  line?  To 
the  same  plane? 

10.  Must  planes  be  parallel  if  they  contain  parallel  lines?  Are  lines 
necessarily  parallel  if  they  are  in  parallel  planes? 

11.  How  many  lines  through  a  point  can  be  perpendicular  to  the  same 
line?     To  the  same  plane? 

12.  How  many  intersecting  planes  can  be  perpendicular  to  the  same 
line?     To  the  same  plane? 

13.  What  are  the  relations  of  the  following  to  each  other:  (1)  lines 
parallel  to  the  same  line;  (2)  planes  parallel  to  the  same  plane;  (3)  lines 
parallel  to  the  same  plane;  (4)  planes  parallel  to  the  same  line;  (5)  lines 
perpendicular  to  the  same  line;  (6)  lines  perpendicular  to  the  same  plane; 
(7)  planes  perpendicular  to  the  same  line;  (8)  planes  perpendicular  to  the 
same  plane? 

14.  If  two  planes  are  perpendicular  to  each  other,  what  lines  in  one 
are  perpendicular  to  the  other? 

15.  Can  the  projection  of  a  straight  line  upon  a  plane  be  a  curve? 
Can  the  projection  of  a  circle  be  a  straight  line?     Explain. 

16.  Can  the  projection  of  a  square  be  a  square?  A  rectangle?  A 
parallelogram?     Explain. 


310 


SOLID   GEOMETRY 


GENERAL   EXERCISES 
COMPUTATION 

1.  Two  fixed  points,  A  and  B,  are  10  in.  apart.  A  point  P  moves  so 
as  to  keep  8  in.  from  A  and  B.     Find  the  length  of  the  locus  of  P. 

2.  Two  fixed  points,  A  and  B,  are  12  in.  apart.  A  point  P  moves  so 
as  to  keep  10  in.  from  A  and  4  in.  from  B.  Find  the  radius  of  the  circle 
generated.  Ans.     2.5  in. 

3.  A  20-foot  pole  is  placed  obliquely  in  a  river.  One  end  of  the  pole 
is  on  the  bottom  of  the  river  and  the  other  end  is  3  ft.  above  the  surface. 
Find  the  depth  of  the  river  if  the  length  of  the  part 
of  the  pole  under  water  is  15  ft. 

4.  Plane  Q  is  perpendicular  to  plane  P.     Find  thex 
shortest  line  AED  that  can  be  drawn  from  a  point  A  in 
Q  to  a  point  D  in  P  if  AB  =  5  ft.,  CD  =  10  ft.,  and  BC  = 
90.  ft. 

Suggestion.     Consider  the  figure  when  Q  is  turned 
into  the  same  plane  as  P. 

5.  The  room  shown  in  the  figure  has  a  length   B< 
of  40  ft.,  a  width  of  20  ft.,  and  a  height  of  12  ft. 
Find  the  length  of  the  shortest  line  in  the  walls  of 

A 


the  room  from  B  to  H. 

6.    The  sum  of  the  face  angles  of  a  tri 


/\ 

/ 

X-   - 

F 

,'V 

edral  angle  is   148°.     What  is  the  greatest 
value  a  face  angle  can  have? 

7.  In  the  figure  of  a  hammock  support 

with  the  dimensions  as  given,  find  the  length  a"--- , Q^'d* 

oiABHAB  =  CB  =  DE  =  FE.  U 

8.  The  roof  of  a  house  makes  an  angle  of  30°  with 

A 
the  plane  of  the  plates.     The  roof  is   18  ft.  by  30  ft. 

Find  the  area  of  its  projection  upon  the  plane  of  the 

plates. 

9.  In  the  figure  of  a  cube  an  edge  is  18  in.  and  BE 
is  10  in.     Find  the  area  of  ABCD. 

10.    Having  given  the  dimensions  as  shown  in 
the  figure,  find  the  area  of  section  S  to  two  decimal  ®* 
places.     All  the  face  planes  that  meet  are  perpen- 
dicular to  each  other. 


GENERAL  EXERCISES 


311 


11.  The  figure  shows  the  plan  for  a  half 
pitch  roof.  The  dimensions  are  as  given  and 
the  rafters  are  to  be  placed  1  ft.  6  in.  from 
center  to  center.  Find  the  lengths  of  the  hip 
rafters,  common  rafters,  and  jack  rafters.  AB 
is  a  hip  rafter,  CD  is  a  common  rafter,  and  EF 
is  a  jack  rafter. 


ko' 


\.             22'             / 

SD    Dech           1 



a 

\ 

A     E    C 


THEOREMS  AND  PROBLEMS 

1.  If  a  series  of  parallel  planes  intercept  equal  segments  on  one  straight 
line,  they  intercept  equal  segments  on  any  other  straight  line  that  they 
intersect. 

2.  If  a  straight  line  and  a  plane  are  each  perpendicular  to  the  same 
straight  line,  they  are  parallel. 

3.  If  two  planes  are  perpendicular  to  each  other,  a  line  perpendicular 
to  one  and  not  in  the  other  is  parallel  to  the  other. 

4.  Prove  that  the  sides  of  an  isosceles  triangle  make  equal  angles 
wfth  any  plane  in  which  the  base  lies. 

5.  From  a  point  within  a  triedral  angle  perpendiculars  are  drawn  to 
the  three  faces.  These  perpendiculars  are  the  edges  of  a  second  triedral 
angle.  Prove  that  the  face  angles  of  one  are  supplementary  to  the  diedral 
angles  of  the  other. 

6.  Three  planes,  M,  N,  and  Q,  each  perpendicular  to  the  other  two 
pass  through  a  common  point.  Show  that  a  point  5  in.  from  M,  8  in.  from 
N,  and  10  in.  from  Q  may  be  in  any  one  of  eight  positions. 

7.  Given  two  non-coplanar  lines,  to  construct  a  plane  upon  which  the 
projections  of  the  two  lines  will  be  parallel. 

8.  A  ray  of  light  from  the  source  L  is  reflected 
by  the  mirror  M  and  enters  the  eye  at  E.  The  path 
travelled  by  the  light  is  the  shortest  possible  path 
from  L  to  the  mirror  and  then  to  E.  Show  how  to 
determine  where  the  ray  of  light  strikes  the  mirror. 

9.  Three  non-coplanar  lines  meet  at  a  point.  Construct  a  line  that 
will  make  equal  angles  with  the  three  lines. 

10.  Given  a  polyedral  angle  with  four  faces,  to  construct  a  plane  that 
will  intersect  the  four  faces  so  that  the  intersection  will  be  a  parallelogram. 

Suggestion.  Pass  a  plane  parallel  to  one  face  and  intersecting  the 
opposite  face.  Determine  a  line  in  the  first  face  equal  to  this  intersection 
and  parallel  to  it.  These  two  fines  are  parallel  and  hence  determine  a 
parallelogram. 


312 


SOLID  GEOMETRY 


11.  Prove  that  if  a  line  is  parallel  to  one  plane  and  perpendicular  to 
another,  the  two  planes  are  perpendicular  to 

each  other. 

12.  If  a  plane  be  passed  through  a  diagonal 
of  a  parallelogram,  the  perpendiculars  to  it 
from  the  extremities  of  the  other  diagonal 
are  equal. 

Plane  MN  passes  through  diagonal  BD  of 
parallelogram  ABCD.  Prove  perpendiculars 
AP  and  CQ  are  equal. 

13.  Describe  how  to  locate  a  point  that  is  4  in.  from  face  of  a  triedral 
angle,  5  in.  from  another  face,  and  3  in.  from  the  third  face. 

14.  If  a  series  of  parallel  planes  cut  all  the  edges  of  a  triedral  angle, 
the  intersections  of  the  planes  with  the  faces 
form  similar  triangles. 

15.  If  from  P,  any  point  within  the  diedral 
angle  A-BC-D,  PM  and  PN  are  drawn  perpen- 
dicular to  the  faces  BD  and  AC  respectively, 
and  MS  is  drawn  perpendicular  to  AC,  then 
NS  is  perpendicular  to  BC. 

16.  The  three  planes  bisecting  the  diedral 
angles  of  a  triedral  angle  meet  in  a  line. 

Suggestion.  Show  that  two  of  them  meet  in 
a  line,  and  then  show  that  this  line  lies  in  the  third 
plane. 

Compare  this  exercise  with  Exercise  13,  page  308. 

17.  All  points  within  a  triedral  angle  and  equally 
distant  from  its  three  faces,  lie  in  the  line  of  inter- 
section of  the  planes  that  bisect  the  diedral  angles. 

18.  The  planes  through  the  bisectors  of  the  face  angle  of  a  triedral 
angle  and  perpendicular  to  the  planes  of  the  respective  faces,  meet  in  a  line. 


CHAPTER  VII 
POLYEDRONS,  PRISMS,   CYLINDERS 

613.  In  the  present  and  following  chapters  will  be  consid- 
ered some  of  the  solids  most  commonly  observed  in  nature, 
and  very  frequently  used  in  architecture,  engineering,  and  the 
arts.  Besides  finding  the  areas  and  volumes  of  these  solids  it 
will  be  necessary  to  investigate  the  relations  of  their  parts, 
and  to  study  the  plane  figures  formed  when  the  solids  are 
cut  by  planes. 

614.  Polyedrons.  A  polyedron  is  a  solid  entirely  bounded 
by  planes. 


The  intersections  of  the  bounding  planes  are  the  edges  of 
the  polyedron.  The  points  in  which  the  edges  intersect  are 
the  vertices.  The  polygons  bounded  by  the  edges  are  the 
faces.  The  faces  taken  together  make  up  the  surface,  and 
the  area  of  this  surface  is  the  area  of  the  polyedron.  The 
:  amount  of  space  enclosed  by  the  surface  is  the  volume  of  the 
•  polyedron. 

615.  Sections.  The  plane  figure  formed 
on  a  plane  passing  through  a  solid,  and 
bounded  by  the  intersections  of  the  plane 
with  the  surface  of  the  solid  is  called  a  sec- 
tion of  the  solid.  It  is  evident  that  the  sec- 
tion of  a  solid  is  a  closed  figure  (§  146),  and  that  the  section 
of  a  polyedron  is  a  polygon. 

313 


314 


SOLID  GEOMETRY 


616.  Convex  Polyedrons.  A  polyedron  is  convex  if  every 
section  of  it  is  a  convex  polygon.     Otherwise  it  is  concave. 

Only  convex  polyedrons  are  considered  unless  otherwise  stated. 

617.  Prismatic  and  Cylindrical  Surfaces.  A  moving  straight 
line  that  always  remains  parallel  to  its  original  position,  and 
intersects  a  fixed  straight  line,  generates  a  plane.     Why? 


N 


Plane 


Prismatic  Surface 


Cylindrical  Surface 


A  moving  straight  line  that  always  remains  parallel  to  its 
original  position,  and  intersects  a  broken  line  not  coplanar 
with  it,  generates  a  prismatic  surface. 

A  moving  straight  line  that  always  remains  parallel  to  its 
original  position,  and  intersects  a  plane  curved  line  not  coplanar 
with  it,  generates  a  cylindrical  surface. 

AB  is  the  original  position  of  the  moving  line,  and  MN  is  the  fixed  line, 

618.  The  moving  line  is  called  the  generatrix,  and 
the  fixed  line  or  curve  the  directrix. 

The  generatrix  in  any  position  is  called  an  element 
of  the  surface  generated. 

619.  Closed  Surface.     If  the  directrix  is  a  closed 
line,  the  prismatic  or  cylindrical  surface  is  closed. 

620.  Sections  of  Prismatic  and  Cylindrical  Surfaces.  A 
plane  cutting  all  the  elements  of  a  closed  prismatic  or  cylin- 
drical surface  cuts  the  surface  in  a  closed  line. 
The  figure  bounded  by  this  closed  line  is  a  sec- 
tion. If  the  cutting  plane  is  perpendicular  to  an 
element,  the  section  is  a  right  section;  if  not,  it 
is  an  oblique  section.  Two  sections  made  by 
parallel  planes  are  parallel  sections. 

In  the  figure,  ABODE  is  a  right  section. 


PRISMS  AND  CYLINDERS 


315 


PRISMS  AND   CYLINDERS 

621.  A  prism  is  the   solid  formed  by  a  closed  prismatic 
surface  and  two  parallel  cutting  planes. 

622.  A  cylinder  is  the  solid  formed  by  a  closed  cylindrical 
surface  and  two   parallel   cutting   planes. 

623*  The  sections  formed  by  the  cut- 
ting planes  are  called  the  bases  of  the 
prism  or  cylinder.  It  follows  from  §  620 
that  the  bases  of  a  prism  are  polygons. 

624.  The  polygons  formed  on  the  pris- 
matic surface  between  the  bases  are  called  the  lateral  faces. 
Intersections  of  the  lateral  faces  are  lateral  edges,  and  intersec- 
tions of  the  lateral  faces  with  the  bases  are  base  edges. 

625.  The  altitude  of  a  prism,   or  cylinder,  is  the  perpen- 
dicular distance  between  its  bases. 

626.  The  lateral  area  of  a  prism,  or  cyl- 
inder, is  the  prismatic,  or  cylindrical,  surface 
between  the  bases.  The  total  area  is  the 
lateral  area  together  with  the  areas  of  the 
bases. 

In  the  figure  read  the  bases,  faces,  lateral  edges, 
base  edges,  altitude. 

627.  The  following  facts  about  prisms  and  cylinders  are 
readily  proved  by  the  definitions  and  previous  theorems: 

(1)  The  lateral  edges  of  a  prism,  or  the  elements  of  a  cylinder, 
are  equal. 

(2)  The  lateral  faces  of  a  prism  are  parallelograms. 

(3)  The  right  section  of  a  prism,  or  cylinder,  is  perpendicular 
to  all  the  lateral  edges,  or  elements. 

(4)  The  section  of  a  prism  made  by  a  plane  parallel  to  a  lateral 
edge  is  a  parallelogram. 

(5)  The  section  of  a  cylinder  made  by  a  plane  containing  two 
elements  is  a  parallelogram. 


316 


SOLID  GEOMETRY 


628.  Theorem.    Parallel  sections  of  a  prism  are  congruent 
Given  the  prism  MN,  and  the  parallel 

sections  AD  and  A'D'. 

To  prove  polygon  AD  =  polygon  A'D'. 

Proof.  AB II  A'B'.  Why? 

Hence  AB  =  A'B'.  §  158 

Similarly  BC  =  B'C,   CD  =  C'D', 
DE  =  D'E',  EA=E'A'. 

Also  ZEAB  =  E'A'B',ZABC  =  ZA'B'C, 
etc.  §  567 

The  two  polygons  AD  and  A'D'  are  then  mutually  equilateral 
and  mutually  equiangular.  They  then  can  be  proved  congruent 
by  superposition. 

.\  polygon  AD  =  polygon  A'D'. 

629.  Theorem.     Parallel  sections  of  a  cylinder  are  congruent. 
Given  the  cylinder  MN,  and  the  parallel  sections  P  and  Q 

made  by  the  planes. 

To  prove  P^Q. 

Proof.  Take  two  points  A  and  B  on  the 
perimeter  of  P,  and  let  C  be  any  third  point 
of  that  perimeter. 

Draw  the  elements  through  A,  B,  and  C 
to  the  points  A',  B',  and  C  respectively  in 
the  perimeter  of  Q;  and  draw  AB,  BC,  CA, 
A'B',  B'C,  and  CA'. 

Then  AABC^AA'B'C.  Why? 

And,  if  P  is  superposed  on  Q  with  AB  coinciding  with  A'B',  C 
will  fall  upon  C.  Why? 

But  C  is  any  point  in  the  perimeter  of  P. 

Hence  every  point  in  the  perimeter  of  P  will  fall  upon  a  point 
of  Q,  and  conversely. 

.\P^Q. 

630.  Theorem.  Every  section  of  a  prism,  or  cylinder,  parallel 
to  the  bases  is  congruent  to  them. 


xzn 


PRISMS  AND  CYLINDERS  317 

631.  A  right  prism,   or  cylinder,   is  one  whose  bases  are 
right  sections. 

632.  An  oblique  prism,  or  cylinder,  is  one  whose  bases  are 
oblique  sections. 

633.  A  prism  is  triangular,  quadrangular,  pentagonal,  etc., 
according  as  its  bases  have  three,  four,  five,  etc.,  sides. 


634.  A  regular  prism  is  a  right  prism  whose  bases  are  reg- 
ular polygons. 

635.  A  circular  cylinder  is  a  cylinder  whose  bases  are 
enclosed  by  circles.  If  the  bases  are  also  right  sections,  the 
cylinder  is  a  right  circular  cylinder. 

636.  The  line  joining  the  centers  of  the  bases  of  a  circular 
cylinder  is  called  its  axis.  The  radius  of  the  base  of  a  circu- 
lar cylinder  is  called  the  radius  of  the  cylinder. 

637.  Theorem.  A  lateral  edge  of  a  right  prism,  or  an  ele- 
ment of  a  right  cylinder,  is  equal  to  the  altitude,         §§  620,  563. 

638.  Theorem.  The  lateral  faces  of  a  right  prism  are  rec- 
tangles.   Apply  §  627  (2). 

639.  Theorem.  //  a  rectangle  revolves  about 
me  of  its  sides  as  an  axis,  it  generates  a  right 
circular  cylinder;  and  conversely. 

Show  that  the  side  opposite  the  axis  remains  parallel  to 
ts  original  position  and  moves  along  a  curve.  Also  show 
-hat  the  other  two  sides  generate  parallel  bases  that  are 
circular,  and  are  perpendicular  to  the  elements. 

640.  Cylinder  of  Revolution.  Since  a  right  circular  cylin- 
ler  may  be  generated  by  revolving  a  rectangle  about  one  of 
ts  sides  as  an  axis,  it  is  often  called  a  cylinder  of  revolution. 


318  SOLID  GEOMETRY 

641.  Plane  Tangent  to  a  Cylinder.  A  plane  is  tangent  to  a 
cylinder  if  it  touches  the  cylindrical  surface  in  one  element 
only, 

642.  Theorem.  If  a  plane  is  tangent  to  a  circular  cylinder, 
its  intersection  with  the  plane  of  a  base  is  tangent  to  that  base. 

Given  the  cylinder  OO'  and  the  tangent  _^ 

plane  P,  meeting  the  cylinder  in  the  ele-  C 

ment  AB,  and  intersecting  the  plane  of 
base  0  in  DE. 

To  prove  that  DE  is  tangent  to  base  0.  Jj| 

Proof.     DE  has  one  point  in  common        /    Im, 
with  00.  Why?      L   m 

Also  DE  has  only  one  point  in  common 
with  OO,  for  if  it  had  another  point  as  C  in  common  with  OO, 
plane  P  would  have  a  point  outside  the  element  AB  in  com- 
mon with  the  cylinder,  which  is  impossible.  Why? 
DE  also  lies  in  the  plane  of  OO.  Why? 
.-.  DE  is  tangent  to  OO.  §  285 

643.  Theorem.  The  plane  determined  by  a  tangent  to  a  base 
of  a  circular  cylinder  and  the  element  through  the  point  of  contact 
is  tangent  to  the  cylinder. 


, 


EXERCISES 

1.  The  bases  of  prisms,  or  cylinders,  are  congruent. 

2.  Would  the  theorem  of  §  628  be  true  if  the  sections  were  concave' 

3.  A  line  drawn  parallel  to  the  elements  of  a  circular  cylinder  from  the 
center  of  one  base,  intersects  the  other  base  at  its  center. 

4.  The  axis  of  a  circular  cylinder  is  parallel  to  its  elements. 

5.  The  axis  of  a  circular  cylinder  passes  through  the  center  of  all  sec-: 
tions  parallel  to  the  bases. 

6.  Every  lateral  edge  of  a  prism  is  parallel  to  the  plane  determined  by 
any  other  two  lateral  edges. 

7.  What  is  the  locus  of  all  points  equidistant  from  a  given  straight  line? 

8.  Show  that  the  plane  determined  by  an  element  of  a  circular  cylinder 
and  the  center  of  one  base  will  contain  the  axis. 


AREAS  OF  PRISMS  AND  CYLINDERS 


319 


AREAS   OF  PRISMS  AND   CYLINDERS 

644.    Theorem.     The  lateral  area  of  a  prism  is  equal  to  the 
product  of  the  perimeter  of  a  right  section  and  a  lateral  edge. 

Given  the  prism  MP',  and  its  right  section  ABCDE. 
To  prove  S  =  pe,  where  £  denotes  lateral 
area,  p  the  perimeter  of  a  right  section, 
and  e  a  lateral  edge. 
Proof.     Area  of  OMN'  =  e-AB.      §  355 
Similarly  area  of  ON0'  =  e-BC, 

etc. 
Since  the  lateral  area  is  the  sum  of  the  M 
faces,  adding  these  gives: 

S  =  e(AB+BC+--EA), 

But  AB+BC-\ EA=p. 

r.S  =  pe. 


Why? 
Why? 
Why? 


645.  Theorem.     The  lateral  area  of  a  right  prism  is  equal  to 
the  product  of  the  perimeter  of  its  base  and  its  altitude. 

That  is,  S  =  ph,  where  S  denotes  lateral  area,  p  perimeter  of 
base,  and  h  altitude. 

646.  Definitions.  An  inscribed 
prism  is  a  prism  whose  bases  are 
respectively  inscribed  in  the  bases  of 
a  cylinder.  The  cylinder  is  then  said 
to  be  circumscribed  about  the  prism. 

A  circumscribed  prism  is  a  prism 
whose  bases  are  respectively  circum- 
scribed about  the  bases  of  a  cylinder, 
said  to  be  inscribed  in  the  prism. 


The  cylinder  is  then 


647.  It  is  evident  that  the  lateral  edges  of  an  inscribed 
prism  are  elements  of  the  cylinder;  and  that  the  lateral  faces 
of  the  circumscribed  prism  are  tangent  to  the  cylinder  by 
§  641, 


320 


SOLID  GEOMETRY 


648.  Area  of  Cylinder.  As  in  §480,  with  the  circle,  th 
general  idea  of  the  measurement  of  a  plane  area  cannot  b( 
used  here,  for  it  is  impossible  to  apply  a  plane  unit  of  area  tc 
a  cylindrical  surface  and  thus  find  its  numerical  measure. 

Practically,  however, 
the  lateral  area  of  a  right 
circular  cylinder  can  be 
found  by  considering 
the  surface  as  straight- 
ened out  and  measured 


S  «  2Tf  i 


as  a  rectangle  having  a  length  equal  to  the  circumference  of  the; 
base,  and  with  a  width  equal  to  the  altitude  of  the  cylinder. 

It  remains  to  devise  a  method  for  measuring  the  surface  o 
a  cylinder,  which  is  accurate  and  also  geometric. 

649.  In  the  circular  cylinder  of  base  AB,  a  prism  having  i 
regular  polygon  for  base  is  inscribed,  and  the  num- 
ber of  sides  is  doubled  indefinitely  according  to 
§  463.  The  area  of  the  cylinder  is  a  constant  and 
that  of  the  prism  is  a  variable.  It  is  evident  that 
the  area  of  each  prism  is  greater  than  the  area 
of  the  prism  of  half  the  number  of  faces. 

It  is  also  evident  that,  at  all  times,  the  area  of  the  cylinde 
is  greater  than  the  area  of  each  successive  prism.     By  doub- 
ling the  number  of  faces  of  the  prism  indefinitely,  the  are; 
continually  increases,   and  can  be  made  to   differ  from  th* 
area  of  the  cylinder  by  less  than  any  assigned  value,  that  is, 
can  be  made  just  as  nearly  equal  to  the  area  of  a  cylinder  as  desired^ 

Similarly,  by  circumscribing  a  prism  having  a 
regular  polygon  for  base  about  a  circular  cylinder 
and  doubling  the  number  of  faces  indefinitely,  the 
area  continually  decreases,  and  can  be  made  to  dif- 
fer from  the  area  of  the  cylinder  by  less  than  any 
assigned  value,  that  is,  it  can  be  made  just  as  nearly  equal 
the  area  of  the  cylinder  as  desired. 


AREAS  OF  PRISMS  AND  CYLINDERS  321 

650.  Volume.  Further,  it  is  evident  that  the  method  of 
the  preceding  paragraphs  can  be  as  readily  applied  to  the 
volume  of  a  cylinder  as  to  its  area.  It  is  thus  seen  that  the 
volume  of  an  inscribed  or  circumscribed  prism,  having  a 
regular  polygon  for  base,  as  the  number  of  faces  is  indefin- 
itely doubled,  can  be  made  just  as  nearly  equal  to  the  volume  of 
the  cylinder  as  desired. 

651.  From  the  preceding  discussion,  the  following  theorem 
may  be  considered  established: 

Theorem.  If  prisms  whose  bases  are  regular  polygons  are 
inscribed  in  and  circumscribed  about  a  circular  cylinder,  and  if 
the  number  of  faces  of  the  prisms  is  indefinitely  doubled, 

(1)  The  perimeter  of  a  right  section  of  the  cylinder  is  the 
common  limit  of  the  perimeters  of  right  sections  of  the  prisms. 

(2)  The  lateral  area  of  the  cylinder  is  the  common  limit  of 
the  lateral  areas  of  the  prisms. 

(3)  The  volume  of  the  cylinder  is  the  common  limit  of  the 
volumes  of  the  prisms. 

652.  Theorem.  The  lateral  area  of  a  circular  cylinder  is  equal 
to  the  product  of  the  perimeter  of  a  right  section  and  the  length  of 
an  element. 

Given  a  circular  cylinder. 

To  prove  S  =  pe,  where  8  denotes  lateral  area, 
p  the  perimeter  of  a  right  section,  and  e  the 
length  of  an  element. 

Proof.     Inscribe  a  prism,  whose  base  is  a  reg- 
ular polygon,  in  the  cylinder,  and  let  S'  denote  its  lateral  area 
and  p'  the  perimeter  of  a  right  section. 

By  doubling  indefinitely  the  number  of  faces  of  the  prism, 
S'->,S,  and  p'->p.         §  652  (1)  and  (2) 

Then  p'e-^pe.  §  485  (2) 

But      S'  =  p'e,  being  variables  that  are  always  equal.     §  644 

,',S  =  pe,  §485(1) 


322 


SOLID  GEOMETRY 


653.  Theorem.  The  lateral  area  of  aright  circular  cylinder 
is  equal  to  the  product  of  its  altitude  and  the  circumference  of  its 
base. 

That  is,  S  =  ch}  where  S  denotes  lateral  area,  c  circumference 
of  base,  and  h  altitude. 

EXERCISES 

1.  Show  that  the  lateral  area  and  the  total  area  of  a  right  circular 
cylinder  of  altitude  h  and  radius  r,  are  given  by  the  formulas: 

S  =  2Wrh,   T  =  2Trh  +  2irr2  =  2TTr{h+r). 

In  the  following  use  the  notation  of  Exercise  1. 

2.  Given  r  =  8  and  h  =  10;  find  S  and  T. 

3.  Given  r  =  10  and  S  =  200;  find  h  and  T. 

4.  Given  h  =  16  and  S  =  256;  find  r  and  T. 

5.  Given  r  =  12  and  T  =  4800;  find  h  and  S. 

6.  Given  ft  =  25  and  T  =  4500;  find  r  and  S. 

7.  Given  S  -  2000  and  T  =  2200;  find  r  and  ft. 

8.  Solve  the  formula  T  =  27Tr/i+27rr2,  (1)  for  r  in  terms  of  h  and  T,  (2) 
for  /i  in  terms  of  r  and  T7. 

9.  From  the  formulas  of  Exercise  1,  find  r  in  terms  of  S  and  T. 

10.  A  peck  measure  made  of  sheet  iron  has  a  diameter  of  8  in.  and  a 
depth  of  10.7  in.     Find  the  number  of  square  inches  of  sheet  iron  in  it. 

11.  Find  the  area  (no  cover)  of  a  wash  boiler  , 
if  the  bottom  is  in  the  form  of  a  rectangle  with  a 
semicircle  at  each  end.     The  rectangle  is  10  in. 
by  14  in.,  and  the  depth  of  the  boiler  is  16  in. 

12.  A  steam  boiler  has  a  diameter  of  72  in., 
is  18  ft.  long,  and  contains  70  tubes  each  having  ' 
a  diameter  of  4  in.  extending  lengthwise  of  the 
boiler.     Find  the  heating  surface  of  the  boiler,  using  " one-half"  in  the 
rule  below.  Arts.     1505  sq.  ft. 

Rule.  In  finding  the  heating  surface  of  a  horizontal  boiler,  it  is  cus- 
tomary to  take  one-half  to  two-thirds  the  lateral  area  of  the  shell,  the 
lateral  area  of  the  tubes,  one-half  to  two-thirds  the  area  of  the  ends  of  the 
boiler,  and  subtract  the  area  of  both  ends  of  the  tubes. 

13.  Find  the  heating  surface  of  a  horizontal  tubular  boiler  12  ft.  long, 
5  ft.  in  diameter,  and  having  52  tubes  2\  in.  in  diameter.  Use  "  two- 
thirds"  in  the  preceding  rule.  Ans.     556.7  sq.  ft. 


AREAS  OF  PRISMS  AND  CYLINDERS  323 

14.  The  base  of  a  right  prism  is  a  triangle  whose  sides  are  12  ft.,  15  ft., 
and  17  ft.,  and  its  altitude  is  8§  ft.     Find  its  lateral  area  and  total  area. 

15.  Find  the  lateral  area  of  a  regular  hexagonal  prism  each  side  of 
whose  base  is  4  in.  and  whose  altitude  is  16  in.     Find  the  total  area. 

16.  Derive  the  formula  S  =0.2QlSdl,  where  8  is  the  area  of  the  surface 
of  a  cylindrical  pipe  in  square  feet,  d  the  diameter  in  inches,  and  I  the 
length  in  feet. 

17.  Six  lines  of  steam  pipes  of  diameter  2  in.  extend  along  one  side  and 
an  end  of  a  room  40  ft.  by  30  ft.  Find  the  number  of  square  feet  of  heat- 
ing surface. 

18.  Small  metal  boxes  used  for  various 
purposes  are  made  from  sheet  metal  as 
follows:  Blanks  of  the  proper  shape  are 
first  cut  from  the  sheet  metal.  These 
are  then  pressed  into  the  required  form 
in  a  die.  In  computing  the  size  of  the 
blank,  it  is  assumed  that  it  has  the  same  area  as  the  finished  article. 

Find  the  diameter  of  the  blank  for  the  cover  of  a  tin  pail  for  holding 
lard,  if  the  cover  has  a  diameter  of  6§  in.  and  the  flange  is  -|  in. 

19.  A  cylindrical  metal  box  for  holding  paper  fasteners  is  If  in.  in 
diameter  and  l|  in.  deep,  and  the  cover  has  a  J-inch  flange.  Find  the 
diameter  of  the  blanks  for  box  and  cover.  Ans.     3.68  in.;  2.56  in. 

20.  What  is  the  locus  of  points  12  in.  distant  from  two  parallel  lines 
18  in.  apart? 

21.  Holes  for  rivets  are  often  punched  in  metal  plates.  The  pressure 
required  for  such  work,  when  the  material  is  wrought  iron,  is  55,000  pounds 
per  square  inch  of  the  cut  surface  in.  For  example,  a  hole  having  a  circum- 
ference of  2  in.  punched  in  a  J-in.  plate  would  require  a  pressure  of 
2  X  J  X  55,000  lb.,  that  is,  the  area  of  the  cylindrical  surface  sheared  off 
times  55,000  lb. 

Find  the  pressure  required  to  punch  a  J-in.  round  hole  through  a  piece 
of  sheet  iron  ^--in.  thick. 

22.  Find  the  pressure  necessary  to  punch  a  ^-inch  round  hole  through 
a  boiler  plate  f  in.  thick  if  60,000  pounds  pressure  is  required  per  square 
inch  of  surface  cut. 

23.  A  rectangular  box  is  to  be  made  of  sheet  steel  -^  in.  thick.  Find 
the  number  of  pounds  the  punch  press  must  strike  in  order  to  cut  out  the 
blank  if  it  is  rectangular  5f  in.  by  2f  in.,  has  squares  -§-  in.  on  a  side  cut 
from  each  corner,  and  has  two  holes  y^-  in.  in  diameter  punched  in  the 
bottom.     Use  60,000  pounds  per  square  inch,  Ans.     30,297  lb. 


324 


SOLID   GEOMETRY 


CONGRUENT   AND   EQUIVALENT   SOLIDS 

654.  Congruent  Solids.     Two  solids  are  said  to  be  congru- 
ent if  they  can  be  made  to  coincide  completely  in  all  their  parts, 

655.  Equivalent  solids  are  those  having  the  same  volume. 

656.  Theorem.     Two    prisms   are    congruent   if  three  face, 
including  a  triedral  angle  of  one  are  congruent  respectively  to 
three  faces  including  a  triedral  angle  of  the  other,  and  are  arranged 
in  the  same  order. 


Given  prism  AI  and  AT,  with  face  AJ  =  face  A'J',  face 
AG^face  A'G',  and  face  AD^face  A'D'. 

To  prove  prism  AI  =  AT. 

Proof.        ZBAF  =  ZB'A'F',  ZFAE  =  ZF'A'E', 
and  ZBAE  =  ZB'A'E'.  Why? 

Then  triedral  ZA  ^triedral  ZA'.  §  610 

Place  the  prism  A' I'  upon  prism  A I  so  that  triedral  A  A'  shall 
coincide  with  its  congruent  triedral  ZA. 

Then  face  A'J'  coincides  with  its  congruent  face  A  J;  A'G' 
with  its  congruent  face  AG;  and  A'D'  with  its  congruent  face  AD. 

And  points  F'$  G',  and  J'  fall  upon  F,  G,  and  J  respectively. 

Further  C'H'  coincides  with  CH.  Why? 

Hence  H'  falls  upon  H. 

Similarly  I'  falls  upon  I. 

.\  prism  AI  =  prism  AT.  §  654 

657.    Theorem.     Two  right  prisms,  or  two  right  cylinders,  are 
congruent  if  they  have  congruent  bases  and  equal  altitudes. 


CONGRUENT   AND   EQUIVALENT   SOLIDS 


325 


658.  Truncated  Prism.  A  portion  of  a  prism  included  be- 
tween the  base  and  a  section  oblique  to  the  base  is  called  a 
truncated  prism. 

659.  Theorem.  Two  truncated  prisms  are  con- 
gruent if  three  faces  including  a  triedral  angle  of  one 
are  congruent  respectively  to  the  three  faces  including 
a  triedral  angle  of  the  other,  and  are  arranged  in  the 
same  order. 

660.  Theorem.  An  oblique  prism  is  equivalent  to  a  right 
prism  whose  base  is  a  right  section  of  the  oblique  prism,  and 
whose  altitude  is  a  lateral  edge  of  the  oblique  prism. 

Given  the  oblique  prism  AD';  also 
the  right  prism  GJ'  whose  base  GJ  is  a 
right  section  of  the  prism  AD',  and 
whose  altitude  is  equal  to  a  lateral  edge 
A  A'  of  prism  AD'. 

To    prove    prism    AD' =  prism    GJ'. 

Outline  of  proof.  Show  that  AG  = 
A'G',  BH  =  B'H',  etc.;  that  the  angles 
of  face  AH  equal  the  angles  of  face 
A'H',  and  that  face  AH ^f ace  A'H'. 

Likewise  show  face  £I  =  face  B'V ,  and  face  AD  =  face  A'D'. 

Then  prism  AJ  =  prism  A' J'.  §  656 

Prism  AJ+prism  GD' =  prism  A'J'+prism  GD'.  Why? 

.'.  prism  AD'  =  prism  GJ'.  Why? 

EXERCISES 

1.  Compare  the  theorem  of  §  660  with  the  theorem  of  §  354. 

2.  If  a  wooden  beam  has  a  rectangular  right  cross  section,  snow  that 
if  it  is  sawed  lengthwise  along  a  diagonal  plane  the  two 
prisms  formed  are  congruent. 

3.  In  a  truncated  prism  having  a  parallelogram  for  base, 
the  sum  of  two  opposite  lateral  edges  is  equal  to  the  sum  of  a 
the  other  two  opposite  lateral  edges. 

Prove  a-\-c- 2e,  and  b +d  =  2e. 


326 


SOLID   GEOMETRY 


PARALLELEPIPEDS 

661.  A  parallelepiped  is  a  prism  whose  bases  are  parallel- 
ograms. 

662.  A  right  parallelepiped  is  a  parallelepiped  whose  lateral 
edges  are  perpendicular  to  its  bases. 

663.  An  oblique  parallelepiped  is  a  parallelepiped  whose 
lateral  edges  are  oblique  to  its  bases. 

664.  A  rectangular  parallelepiped  is  a  right  parallelepiped 
whose  bases  are  rectangles. 

665.  A  cube  is  a  rectangular  parallelepiped  whose    edges 
are  all  equal. 


Oblique 
Parallelepiped 


Cube 


Rectangular 
Parallelepiped 


666.  The  dimensions  of  a  rectangular  parallelepiped  are 
the  lengths  of  the  three  edges  drawn  from  one  vertex.  These 
dimensions  are  often  called  length,  breadth,  and  height. 

667.  A  diagonal  of  a  parallelepiped  is  the  line  from  any 
vertex  to  a  vertex  not  in  the  same  face. 

668.  The  volume  of  any  solid  is  the  numerical  measure  of 
its  magnitude  in  terms  of  some  unit  of  measure. 

The  unit  of  measure  usually  taken  is  a  cube  one  linear  unit  on  an  edge. 

669.  Ratio  between  Solids.  By  the  ratio  of  one  solid  to 
another  is  meant  the  ratio  of  their  numerical  measures. 

670.  Prove  the  following  facts  about  parallelepipeds: 

'  (1)  All  the  faces  (including  the  bases)  of  a  parallelepiped  are 
parallelograms. 

(2)  A  parallelepiped  is  bounded  by  three  pairs  of  parallel  con- 
gruent parallelograms. 


PARALLELEPIPEDS  327 

(3)  A  parallelepiped  has  three  sets  of  four  equal  edges. 

(4)  Any  two  opposite  faces  of  a  parallelepiped  may  be  taken 
as  bases. 

(5)  All  the  faces  of  a  rectangular  parallelepiped  are  rectangles. 

(6)  All  the  faces  of  a  cube  are  squares. 

671.  Volume  of  Rectangular  Parallelepiped.  By  a  consid- 
eration analogous  to  that  of  §  344  the  reasonableness  of  the 
following  statement  would  be  evident;  here  it  is  accepted 
without  proof. 

The  volume  of  a  rectangular  parallelepiped  is  equal  to  the  pro- 
duct of  its  three  dimensions. 

If  V,  a,  b,  and  h  are  the  numerical  measures  of  the  volume,  length, 

breadth,  and  height,  respectively,  of  any  rectangular  parallelepiped,  then 

the  above  is  stated  in  the  formula 

V  =  abh. 

This  means  that,  if  the  three  dimensions  are 
each  measured  in  terms  of  some  unit  of  length, 
then  the  volume  is  measured  in  terms  of  the 
corresponding  unit  of  volume,  which  is  a  cube 
having  an  edge  one  linear  unit  in  length. 

672.  Theorem.  The  volume  of  a  rectangular  parallelepiped 
is  equal  to  the  product  of  its  base  and  altitude. 

673.  Theorem.  The  volume  of  a  cube  is  equal  to  the  cube  of 
its  edge. 

674.  Theorem.  The  volumes  of  two  rectangular  parallel- 
epipeds are  to  each  other  as  the  products  of  their  three  dimensions. 
V  :  V'  =  abh'.a'b'h'. 

675.  Theorem.  The  volumes  of  two  rectangular  parallel- 
epipeds having  equivalent  bases  are  to  each  other  as  their  altitudes. 

v     abh      ;  r    ■  ;' \     v    h 

bince  —  =— — —  and  ab=ab  ,  then  —  =— . 
V      abh  V      h 

676.  Theorem.  The  volume  of  two  rectangular  parallel- 
epipeds having  equal  altitudes  are  to  each  other  as  their  bases. 

677.  Theorem.  Two  rectangular  parallelepipeds  having 
equivalent  bases  and  equal  altitudes  are  equal  in  volume. 


328  SOLID  GEOMETRY 

EXERCISES 

1.  Find  the  volumes  of  the  following  rectangular  parallelepipeds: 
(1)  13  ft.  by  27  ft.  by  45  ft.  (2)  2  ft.  3  in.  by  1  ft.  7  in.  by  3  ft.  1  in.  (3) 
30  ft.  6  in.  by  41  ft.  6  in.  by  12  ft. 

2.  A  common  brick  is  8  in.  by  4  in.  by  2  in.  Find  the  number  of 
common  brick  in  a  pile  3  ft.  by  6  ft.  by  12  ft. 

3.  How  many  shoe  boxes  each  3  in.  by  4  in.  by  9  in.  can  be  put  in  a 
packing  box  3  ft.  by  3  ft.  4  in.  by  3  ft.  9  in.? 

4.  Find  the  number  of  cubic  yards  of  earth  to  be  excavated  in  digging 
a  cellar  40  ft.  by  26  ft.  by  7  ft. 

5.  If  180  sq.  ft.  of  zinc  are  required  to  line  the  bottom  and  sides  of  a 
cubical  vessel,  how  many  cubic  feet  of  water  will  it  hold? 

6.  A  box  car  that  is  36|  ft.  long  and  8  ft.  wide,  inside  measurements, 
can  be  filled  with  wheat  to  a  height  of  4^  ft.  Find  how  many  bushels  of 
wheat  it  will  hold  if  -f-  cu.  ft.  are  a  bushel. 

7.  Find  the  cost  at  40  cents  a  pound  for  sheet  copper  to  line  the  bot- 
tom and  sides  of  a  cubical  vessel  7  ft.  on  an  edge,  if  the  sheet  copper  weighs 
12  oz.  per  square  foot.     Find  volume.  Ans.     $73.50;  81.45  bbl. 

8.  Find  the  cost  of  laying  a  stone  wall  45  ft.  long,  6  ft.  high,  and  2  ft. 
thick  at  $2.75  a  perch.     Use  22  cu.  ft.  for  one  perch. 

9.  The  edge  of  a  cube  is  10  in.  Find  the  edge  of  a  cube  which  shall 
have  a  volume  twice  as  great.     Eight  times  as  great. 

10.  Show  that  the  edge  and  diagonal  of  a  cube  can  be  used  as  the  two 
sides  of  a  right  triangle  whose  acute  angles  are  30°  and  60°. 

11.  Are  the  diagonals  of  a  rectangular  parallelepiped  perpendicular  to 
each  other?     Are  those  of  a  cube?     Prove. 

12.  The  diagonals  of  a  rectangular  parallelepiped  are  equal. 

13.  The  diagonals  of  a  rectangular  parallelepiped  bisect  each  other. 

14.  Find  the  sum  of  all  the  face  angles  of  a  parallelepiped. 

15.  Find  the  edge  of  a  cube  if  its  volume  is  increased  200  cu.  in.  when 
each  edge  is  increased  2  in. 

16.  The  edges  of  a  rectangular  parallelepiped  are  a,  b,  and  c.  Find 
the  length  of  a  diagonal,  and  the  entire  area  of  the  parallelepiped. 

17.  Test  the  following  rule  if  1  cu.  in.  of  iron  weighs  0.28  pounds:  The 
weight  of  iron  bars  in  pounds  per  foot  of  length  equals  the  width  in  inches 
times  the  thickness  in  inches  times  ^-. 

18.  The  sum  of  the  squares  of  the  four  diagonals  of  any  parallelepiped 
is  equal  to  the  sum  of  the  squares  of  the  twelve  edges.     Apply  Ex.  1 1,  p.  181. 


PARALLELEPIPEDS  329 

678.    Theorem.     The  volume  of  any  parallelepiped  is  equal 
to  the  product  of  its  base  and  altitude.     V  =  Bh. 


Given  X  an  oblique  parallelepiped,  with  area  of  base  AC  —  B, 
and  height  h. 

To  prove  V  =  Bh,  where  V  denotes  volume. 

Proof.  Extend  DC  and  the  other  edges  of  X  that  are  parallel 
to  it. 

On  DC  extended,  take  EF  =  DC. 

Pass  planes  EG  and  FH  through  E  and  F  and  perpendicular 
to  DC,  forming  a  right  parallelepiped  Y,  in  which  EG  is  a  right 
section. 

Then  parallelepiped  X  =  parallelepiped  Y.  §  660 

Extend  IE  and  the  other  edges  of  Y  that  are  parallel  to  it. 

On  IE  extended,  take  JK  =  IE. 

Pass  planes  JL  and  KM  through  J  and  K  and  perpendicular 
to  IE,  forming  a  rectangular  parallelepiped  Z. 

Then  parallelepiped  Y  =  parallelepiped  Z.  §  660 

Therefore  parallelepiped  X  =  parallelepiped  Z.  §  104 

But  volume  of  Z  =  base  JN  X  NM.  §  672 

In  which  NM  =  h.  Why? 

And  UJN  =  n}IF  =  OAC  =  B.  Why? 

/.  V  =  Bh.  Why? 

679.  Theorem.  Parallelepipeds  having  equivalent  bases  and 
equal  altitudes  are  equivalent. 


330  SOLID  GEOMETRY 

680.  Theorem.  Two  parallelepipeds  are  to  each  other  as  the 
products  of  their  bases  and  altitudes. 

681.  Theorem.  Two  parallelepipeds  having  equal  altitudes 
are  to  each  other  as  their  bases. 

682.  Theorem.  Two  parallelepipeds  having  equivalent  bases 
are  to  each  other  as  their  altitudes. 

EXERCISES 

1.  Compare  theorems  of  §§  678-682  with  those  of  §§  355,  358-361. 

2.  Use  the  notation  of  §  678,  and  show  that  in  any  parallelepiped 
V  V 

3.  Find  the  length  of  the  diagonal  of  a  rectangular  parallelepiped 
whose  dimensions  are  30  ft.,  40  ft.,  and  12  ft. 

4.  A  parallelepiped  has  a  base  in  the  form  of  a  rhombus  whose  edges 
and  one  diagonal  are  each  10  in.     Find  the  volume  if  the  altitude  is  8  in. 

5.  A  parallelepiped  has  a  rectangular  base  8  in.  by  15  in.,  and  square 
ends.  Find  its  volume  if  one  of  the  sides  is  a  parallelogram  having  an 
angle  of  60°. 

6.  Find  the  diagonal  of  a  cube  whose  volume  is  512  cu.  in. 

7.  Find  the  dimensions  of  a  rectangular  parallelepiped  having  a  vol- 
ume of  12,960  cu.  in.,  if  the  dimensions  are  in  the  ratio  of  3  :  4  :  5. 

Ans.     18  in.,  24  in.,  30  in. 

8.  Find  the  edge  of  a  cube  whose  surface  and  volume  have  the  same 
numerical  value. 

9.  A  parallelepiped  of  altitude  8  in.  has  the  same  volume  as  a  cube 
that  is  12  in.  on  an  edge.  Find  the  area  of  the  base  and  the  dimensions 
of  the  base  if  it  is  a  parallelogram  having  an  angle  of  60°  and  one  side  equal 
to  18  in. 

10.  Find  the  length  of  the  bar  that  can  be  made  from  1  cu.  ft.  of  steel, 
if  the  bar  has  a  rectangular  cross  section  §  in.  by  1  \  in. 

11.  Find  the  volume  to  the  nearest  .001  cu.  ft.  of  a  rectangular  tank 
17  ft.  2  in.  by  19  ft.  3  in.  by  3  ft.  7  in.  Ans.     1184.142  cu.  ft. 

12.  Show  geometrically  that  (a+6)3=a3+3a26+3a62+63. 

13.  The  total  area  of  a  right  prism  whose  base  is  a  rectangle  is  118  sq. 
ft.,  the  volume  is  70  cu.  ft.,  and  the  altitude  is  7  ft.  Find  dimensions  of 
the  base. 


VOLUMES  OF  PRISMS  AND  CYLINDERS  331 

VOLUMES   OF  PRISMS  AND   CYLINDERS 

683.  Theorem.  The  plane  passed  through  the  two  diagonally 
opposite  edges  of  a  parallelepiped  divides  the  parallelepiped  into 
two  equivalent  triangular  prisms. 


A 

"    jB  / 

Eh 

/   ^^^ 

/ 

Jp7G 

/  • 

r 

'—Jc 

Given  the  parallelepiped  AC,  divided  into  two  triangular 
prisms  A'-ABD  and  C'-BCD  by  the  diagonal  plane  passing 
through  the  edges  BB'  and  DD'. 

To  prove  prism  A'-ABD  =  prism  C'-BCD. 

Proof.  Let  EFGH  be  a  right  section  of  the  parallelepiped, 
cutting  the  diagonal  plane  in  FH. 

Face  AB'  II  face  DC,  and  face  AD'  II  face  EC.  Why? 

Then  EF  II  EG,  and  EH  II  FG.  §  562 

And  EFGH  is  a  parallelogram.  Why? 

FHisa  diagonal  of  OEFGH.  Why? 

Hence  AEFH^AFGH.  §  154 

Prism  A'-ABD  equals  a  right  prism  with  base  EFH  and 
altitude  AA'.  §  660 

Prism  C'-BCD  equals  a  right  prism  with  base  FGH  and 
altitude  AA'.  Why? 

But  these  two  right  prisms  are  congruent  and  so  equivalent. 
.-.  prism  A' -ABC  =  C'-BCD.  §  104 

EXERCISES 

1.  Find  the  edge  of  a  cube  that  is  increased  in  volume  127  cu.  in.  when 
its  edge  is  increased  1  in. 

2.  A  rectangular  solid  with  a  square  base  has  a  volume  of  80  cu.  ft., 
and  a  surface  of  1 12  sq.  ft.     Find  its  dimensions.     Ans.  4  ft.  by  4  ft.  by  5  ft. 


332 


SOLID  GEOMETRY 


684.  Theorem.     The  volume  of  a  triangular  prism  is  equal 
to  the  product  of  its  base  and  altitude.     V  =  Bh. 

Given   the   triangular   prism   A' -ACT). 

To  prove  V  =  Bh,  where  V  denotes  vol- 
ume, B  area  of  base,  and  h  altitude. 

Suggestion.  Complete  the  parallel- 
epiped A' -AGED.  Show  that  prism 
A'-ACD=i  parallelepiped  A'-ACED. 
Apply  §  683. 

685.  Theorem.     The  volume  of  any  prism  is  equal  to  the 
product  of  its  base  and  altitude.     V  =  Bh.  E<  D, 

Given  any  prism  AD'.  J/*CS2^>C' 

To  prove  V  =  Bh,  where  V  denotes  volume,  /  Ua 

B  area  of  base,  and  h  altitude. 

Suggestion.  Show  that  the  prism  can  be 
divided  into  triangular  prisms  by  diagonal 
planes.     Add  the  prisms  thus  formed.  f< 

686.  Theorem.  Prisms  having  equivalent 
bases  and  equal  altitudes  are  equivalent. 

687.  Theorem.     Prisms  having  equivalent  bases  are  to  each 
other  as  their  altitudes. 

688.  Theorem.     Prisms  having  equal  altitudes  are  to  each 
other  as  their  bases. 


EXERCISES 

1.  Find  the  volume  of  a  prism  whose  base  is  a  regular  hexagon  6  in. 
on  a  side,  and  whose  altitude  is  24  in. 

2.  In  a  prism  given  V  =  226  and  B  =  43.6;  find  h. 

3.  The  sides  of  a  right  section  of  a  triangular  prism  are  4  in.,  5  in., 
and  7  in.     Find  the  volume  if  a  lateral  edge  is  16  in. 

4.  The  cost  of  digging  a  ditch,  including  all  expenses  and  profits,  is 
estimated  at  27  cents  a  cubic  yard.  Find  the  cost  of  digging  a  ditch  15  mi. 
long,  10  ft.  wide  at  the  bottom,  20  ft.  at  the  top,  and  6  ft.  deep.  The  cross 
section  is  a  trapezoid  Arts.     $71,280. 


VOLUMES  OF  PRISMS  AND  CYLINDERS 


333 


5.  Show  that  an  oblique  prism  of  wood  may  be  changed  into  a  right 
prism  by  cutting  along  a  right  section  and  interchanging  the  two  parts. 

6.  One  of  the  concrete  pillars  to  support  a  floor  in  a  concrete  building  is 
12  ft.  high  and  has  as  a  cross  section  a  regular  hexagon  8  in.  on  a  side.  Find 
its  weight  if  the  concrete  weighs  138  lb.  per  cubic  foot.        Arts.     1912  lb. 

7.  A  flow  of  300  gallons  per  second  will  supply  water  for  a  stream  of 
what  depth,  if  the  stream  is  4  ft.  wide  and  flows  5  miles  per  hour? 

Ans.     16-H-  in. 

8.  Find  the  number  of  cubic  yards  of 
crushed  rock  to  make  a  road  1  mile  in  length 
and  of  cross  section  as  shown  in  the  figure. 

9.  One  cubic  inch  of  steel  weighs  0.29  lb.  An  I-beam 
has  a  cross  section  as  shown  in  the  figure  and  a  length 
of  22  ft.     Find  its  weight. 

10.  An  iron  casting  shrinks  f  in.  per  linear  foot  in  cool- 
ing down  to  70  degrees  Fahrenheit.  How  many  cubic 
inches  is  the  shrinkage  per  cubic  foot? 

Ans.     53.44  cu.  in. 

11.  How  many  cubic  yards  of  soil  will  it  take  to  fill  in  a  lot  50  ft.  by 
100  ft.  if  it  is  to  be  raised  3  ft.  in  the  rear  and  gradually  sloped  to  the  front 
where  it  is  to  be  1^  ft.  deep?  Ans.     416§  cu.  yd. 

12.  In  a  regular  triangular  prism,  the  edge  of  the  base  and  the  altitude 
are  equal.     Find  these  dimensions  if  the  volume  is  128\/3  cu.  in. 

13.  In  a  regular  triangular  prism,  the  altitude  is  6  in.  more  than  the 
edge  of  the   base.     Find   the   dimensions   if   the 

volume  is  224  \/3  cu.  in. 

14.  Find  the  volume  of  a  regular  hexagonal 
prism  whose  base  has  an  area  of  37.5  sq.  ft.,  and 
whose  altitude  equals  an  edge  of  the  base. 

15.  The  perpendicular  drawn  to  the  lower  base 
of  a  truncated  triangular  prism  from  the  intersection 
of  the  medians  of  the  upper  base,  equals  one  third 
the  sum  of  the  three  lateral  edges. 

16.  The  volume  of  any  oblique  prism  is  equal  to  the  product  of  the  area 
of  a  right  section  by  the  length  of  a  lateral  edge. 

17.  The  volume  of  a  regular  hexagonal  prism  is  30\/3  cu.  in.  and  its 
lateral  area  is  180  sq.  in.     Find  its  altitude  and  base  edge. 

18.  The  volume  of  a  triangular  prism  is  equal  to  the  area  of  a  lateral 
face  times  one-half  the  perpendicular  drawn  to  that  face  from  the  opposite 
edge. 


334  SOLID  GEOMETRY 

689.    Theorem.     The  volume  of  a  circular  cylinder  is  equal 
to  the  product  of  its  base  and  altitude.    V  =  Bh. 

Given  a  circular  cylinder. 

To  prove  that  V  =  Bh,  where  V  denotes  vol- 
ume, B  area  of  base,  and  h  altitude. 

Proof.  Inscribe  a  prism,  whose  base  is  a 
regular  polygon,  in  the  cylinder,  and  let  V 
denote  its  volume  and  B'  the  area  of  its  base. 

By  doubling  indefinitely  the  number  of  faces  of  the  prism, 
V-+V,  and  B'->B.  §§  651  (3),  489 

Then  B'h-^Bh.  §485(2) 

But      V'=B'h,  being  variables  that  are  always  equal.    §  685 
.'.V  =  Bh.  §485(1) 

EXERCISES 

1.  Show  that  the  volume  of  a  hollow  cylinder  with  outer  diameter  D, 
inner  diameter  d,  and  altitude  h,  is  given  by  the  formula, 

V  =  lTTh(D*-d*)  =lirh(D+d)(D-d).     (See  Ex.  8,  p.  249.) 

2.  A  cylindrical  oil  tank  3  ft.  in  diameter  and  10  ft.  long  will  contain 
how  many  gallons?     (1  gal.  =231  cu.  in.)  Arts.     528.8. 

3.  The  outer  diameter  of  a  hollow  cast  iron  shaft  is  18  in.;  and  its 
inner  diameter  is  10  in.  Calculate  its  weight  if  the  length  is  20  ft.  and 
cast  iron  weighs  0.26  lb.  per  cubic  inch. 

4.  A  peck  measure  is  to  have  a  diameter  of  8  in.  How  deep  should 
it  be?     (lbu.=  2150.42  cu.  in.) 

5.  Water  is  flowing  at  the  rate  of  10  miles  per  hour  through  a  pipe  of 
diameter  16  in.  into  a  rectangular  reservoir  197  yd.  long  and  87  yd.  wide. 
Calculate  the  time  in  which  the  surface  of  the  water  in  the  tank  will  be 
raised  3  in.  Arts.     31.38  minutes. 

6.  A  certain  handbook  gives  the  following  "rules  of  thumb"  for  find- 
ing the  volume  in  gallons  of  a  cylindrical  tank: 

(1)  V  =  (diameter  in  feet)2  X  5 J  X  (height  in  feet) . 

(2)  V  =  (diameter  in  feet)2  X  §  (height  in  inches)  less  2%  of  the  product. 
Find  the  per  cent  of  error  for  each  rule. 

7.  Find  the  height  of  a  10-gallon  wash  boiler  whose  base  is  10  in. 
wide  with  semicircular  ends,  the  length  of  the  straight  part  of  the  sides 
being  9J  in.  Arts.     13.5  in 


VOLUMES  OF  PRISMS  AND   CYLINDERS 


335 


8.  A  cylinder  to  cool  lard  is  4  ft.  in  diameter  and  9  ft.  long  and  makes 
four  revolutions  per  minute.  At  each  revolution,  the  hot  lard  cools  upon 
the  surface  to  a  depth  of  j  in.  How  many  pounds  of  lard  will  it  cool  in 
one  hour  if  1  cu.  ft.  of  lard  weighs  56|  pounds? 

9.  If  a  tank  5  ft.  in  diameter  and  10  ft.  deep  holds  10,000  pounds  of 
lard,  what  will  be  the  depth  of  a  tank  of  2000  pounds  capacity  if  its  diam- 
eter is  3  ft.?  If  this  tank  has  a  jacket  around  it  on  the  bottom  and  sides 
3  in.  from  the  surface  of  the  tank,  how  many  gallons  of  water 
will  the  space  between  the  jacket  and  the  tank  hold? 

10.  A  conduit  made  of  concrete  has  a  cross  section  with 
dimensions  as  shown  in  the  figure.  How  many  cubic  yards 
of  concrete  are  used  in  making  one  mile  of  this  conduit? 

11.  The  cylindrical  water  tower  shown  in  the  figure 
is  at  Long  Beach,  N.Y.  Its  diameter  is  34  ft.,  height 
150  ft.,  and  it  is  said  to  have  a  capacity  of  1,020,000 
gallons.     Is  the  capacity  given  correct? 

12.  Does  a  cylindrical  water  tank  42  in.  in  diameter 
and  14  ft.  long  hold  1000  gallons? 

13.  The  following  "rule  of  thumb"  is  used  for 
finding  the  weight  of  round  iron.  The  weight  of  round 
iron  in  pounds  per  foot  equals  the  square  of  the  diam- 
eter in  quarter  inches  divided  by  6.  Find  the  per 
cent  of  error  in  using  the  rule  if  iron  weighs  0.28  lb.  per  cubic  inch. 

14.  How  much  water  will  a  horizontal  steam  boiler  5  ft.  in  diameter 
and  16  ft.  long  with  70  tubes  of  diameter  3  in.  running  lengthwise,  hold  if 
one  third  of  the  volume  is  for  steam?  x,„ 

15.  Find  the  height  of  a  cylindrical  oil  tank  with  a 
diameter  of  16  in.  to  hold  one  barrel.         Ans.     36.2  in. 

16.  A  tool  steel  ring  for  a  steam  cylinder  is  forged 
from  round  stock  3  in.  in  diameter.  Find  the  length  of 
stock  to  make  a  ring  with  the  dimensions  given  in  the  figure. 

17.  The  segment  in  the  figure  is  a  counter-balance  5 \  in.  thick, 
its  weight  if  made  of  cast-iron  weighing  0.26  lb.  per  n 
cubic  inch.                                                  Ans.  228.5  lb. 

18.  Find  the  weight  of  a  hollow  hexagonal  bar  16 
ft.  long  and  weighing  0.28  lb.  per  cubic  inch.  The 
cross  section  is  a  regular  hexagon  l\  in.  on  a  side, 
with  a  circle  1^  in.  in  diameter  at  the  center. 

.    .  Ans.  152j  lb. 


Find 


336 


SOLID  GEOMETRY 


SIMILAR  PRISMS  AND   CYLINDERS 

690.  Definitions.  Two  right  prisms  having  similar  poly- 
gons for  bases,  and  whose  altitudes  are  in  the  same  ratio  as 
two  corresponding  base  edges,  are  similar. 

Two  right  circular  cylinders  are  similar  if  their  altitudes  are 
in  the  same  ratio  as  the  radii  of  their  bases. 

691.  Theorem.  The  volumes  of  two  similar  right  prisms  are 
in  the  same  ratio  as  the  cubes  of  corresponding  base  edges. 


Given  two  similar  right  prisms  P  and  P',  having  altitudes  h 

and  h',  bases  B  and  B\  and  two  corresponding  base  edges  e  and  e' 

~  V     h*     e* 

To  prove  — = — = — . 

17'     j,/3     pn 


V    W 


Proof. 

Hence 

But 
And 

Also 


V  =  Bh,  and  V '■- 

=  B'h' 

V  _Bh  _B 

.  h 

V    B'h'    B' 

'  h'' 

B  _e2 

¥'~e'2' 

h     e 
h'~7 

V  _e2     e_ 
"  V    e'2'e'~ 

e3 
e'3' 

V  _h3 

V    h'*' 

§685 
§108 

§446 

§690 

§111 

§111 


SIMILAR  PRISMS  AND  CYLINDERS 


337 


692.  Theorem.  The  volumes  of  two  similar  right  circular 
cylinders  are  in  the  same  ratio  as  the  cubes  of  their  altitudes,  radii, 
or  diameters. 


Given  two  similar  right  circular  cylinders  C  and  C",  having 
volumes  V  and  V,  bases  B  and  B',  altitudes  h  and  h',  radii  r 
and  r',  and  diameters  d  and  d' . 

V     h*     r3      d3 
To  prove  —  =  —  =  —  = 

r  Iff         L'3        „'Z 


Y'     hfK  r'3    d'3 
The  proof  is  similar  to  the  preceding. 


r 


EXERCISES 

1.  Using  the  notation  of  §  691  for  similar  prisms,  also  S  and  S'  for 
lateral  areas,  and  T  and  T'  for  total  areas,  prove  the  following: 

-----  (2)  ----- 

2.  Using  the  notation  of  §  692  for  similar  cylinders,  also  S,  S',  T,  and 
7"  as  in  Exercise  1,  prove  the  following: 

3.  Find  the  ratios  of  the  volumes,  the  lateral  areas,  and  the  total  areas 
of  two  similar  right  prisms  having  hexagonal  bases  two  of  whose  corres- 
ponding edges  are  2  in.  and  5  in.  respectively. 

4.  Two  similar  right  circular  cylinders  have  radii  of  3  in.  and  7  in. 
respectively.  Find  the  ratio  of  their  volumes.  Of  their  lateral  areas. 
Of  their  total  areas. 

5.  Two  similar  right  circular  cylinders  have  bases  whose  areas  are 
20.25  sq.  in.  and  100  sq.  in.  respectively.     Find  the  ratio  of  their  volumes. 


338  SOLID  GEOMETRY 

6.  The  total  areas  of  two  similar  right  circular  cylinders  are  625  sq.  in. 
and  324  sq.  in.  respectively.  Find  the  diameter  of  the  second  if  the  diam- 
eter of  the  first  is  6j  in. 

7.  A  rectangle  having  dimensions  of  10  in.  and  15  in.  is  revolved,  first 
about  the  side  that  is  10  in.,  and  second,  about  the  side  that  is  15  in. 
Find  the  ratio  of  the  volumes  of  the  two  cylinders  formed. 

8.  The  number  of  feet  of  lumber  in  a  log  is  often  based  upon  a  standard 

log  usually  12  ft.  long  and  24  in.  in  diameter  inside  the  bark  at  the  small 

end.     If  v,  d,  and  I  are  the  volume,  diameter,  and  length  respectively  of 

the  standard  log;  and  V,  D,  and  L,  the  corresponding   measurements  of 

vD2L 
the  log  to  be  measured,  then  V=  .     Derive  this  formula. 


QUESTIONS 

1.  What  is  a  prismatic  surface?  A  cylindrical  surface?  Is  the  direc- 
trix necessarily  closed? 

2.  In  §  617,  suppose  the  directrix  is  coplanar  with  the  generatrix, 
what  is  the  form  of  the  surface  generated? 

3.  What  are  the  formulas  or  rules  for  finding  areas  of  prisms  and 
cylinders?  What  are  the  formulas  or  rules  for  finding  the  volumes  of 
each  of  these  solids? 

4.  Why  is  there  such  a  close  relation  between  theorems  concerning 
prisms  and  cylinders? 

5.  Can  you  find  the  area  of  an  oblique  circular  cylinder?  Of  an 
oblique  prism?     Can  you  find  the  volume  of  each  of  these? 

6.  Trace  the  steps  in  finding  the  volume  of  a  parallelepiped. 

7.  Trace  the  steps  in  finding  the  volume  of  any  prism.     Of  a  cylinder. 

8.  What  effect  does  it  have  upon  the  volume  of  a  prism  or  cylinder  if 
the  base  is  doubled?  If  the  altitude  is  doubled?  If  both  base  and  alti- 
tude are  doubled? 

9.  What  effect  does  it  have  upon  the  lateral  area  of  a  right  circular 
cy Under  if  the  circumference  of  the  base  is  doubled?  If  the  area  of  the 
base  is  doubled? 

10.  Which  do  you  consider  the  more  common  form  in  nature,  the  cyl- 
inder or  the  prism?     Name  some  examples  of  each  form. 

11.  Which  is  the  more  common  of  these  forms  in  buildings  and  archi- 
tecture? In  machinery?  Give  illustrations  of  triangular  prisms,  of  quad- 
rangular prisms,  of  pentagonal,  of  hexagonal,  etc. 


GENERAL  EXERCISES 


339 


GENERAL   EXERCISES 
COMPUTATIONS 

1.  Find  the  number  of  barrels  each  of  the  following  cylindrical  tanks 
will  hold:  (1)  diameter  5  ft.  and  depth  5  ft.,  (2)  diameter  20  ft.  and  depth 
19  ft.     (1  bbl.  =3.211  cu.  ft.) 

2.  Find  the  cost  of  laying  the  stonework,  at  $1.75  per  cubic  yard,  in  two 
abutments  for  a  bridge,  each  abutment  to  be  8  ft.  high,  3 J  ft.  thick,  20  ft. 


V-2&1U 


Top  of 
Pier 


3 


,L 


Ans.     28y  sq.  in. 


long  at  the  bottom,  and  15  ft.  at  the  top. 

3.  Find  the  cost  of  common  brick  in  the  pier  with  a 
cross  section  as  shown  in  the  figure,  and  a  height  of  12 
ft.  6  in.,  at  $7.00  per  thousand.  Count  20  brick  to  1 
cu.  ft.  Ans.     $24.15. 

4.  A  stream  flowing  5  miles  per  hour  must  be  how 
large  in  cross  section  to  supply  water  to  a  depth  of  1  \ 
in.  per  week,  for  160  acres  of  land? 

5.  A  water  tank  in  a  Pullman  car 
has  a  vertical  section  as  shown  in  the 
figure,  and  a  length  of  52  in.  Find  its 
capacity  in  gallons.         Ans.     68.3  gal. 

Consider  the  arc  as  a  part  of  a  cir-_l 
cle  and  apply  formula  (1)  of  §  509. 

6.  The  flanges  at  the  joining  of  two 
ends  of  flanged  steam  pipes  10  in.  in 
inside  diameter  are  bolted  together  by 
14  bolts  f  in.  in  diameter.  If  the  pres- 
sure in  the  pipes  is  200  pounds  per  square 
inch,  find  what  each  bolt  must  hold.  How  much  is  this  per  square  inch 
cross  section  of  the  bolts?  Suppose  that  the  bolts  have  10  pitch  U.S. 
standard  thread.     This  makes  the  root  diameter  0.620  in. 

Ans.     1122  lb.;  3718-  lb. 

7.  Find  the  diameter  of  a  cylindrical  oil  tank  40.5  in.  high  that  is  to 
hold  1  barrel.  Ans.     15j  in. 

8.  Representing  the  dimensions  of  a  rectangular  solid  by  x,  y,  and  2, 
find  their  values  if  when  each  is  increased  2  in.  the  volume  is  increased 
150  cu.  in.,  the  face  of  dimensions  x  and  y  is  increased  18  sq.  in.,  and  the 
total  surface  is  increased  110  sq.  in.  Ans.     3  in.,  4  in.,  and  5  in. 

9.  Two  cubes  whose  edges  differ  by  1  in.  have  volumes  that  differ  by 
397  cu.  in.     Find  the  edges  of  each  cube. 


340 


SOLID  GEOMETRY 


10.  A  rectangular  sheet  of  tin,  12  in. 
by  16  in.,  is  made  into  an  open  box  by  cut-  .. 
ting  out  a  square  from  each  corner  and  83 
turning  up  the  sides.     Find  the  size  of  the 
square  cut  out  if  the  volume  of  the  box  is 
180  cu.  in. 

11.  A  rectangular  piece  of  tin  a  inches  longer  than  it  is  wide  is  made 
into  an  open  box,  containing  c  cubic  inches,  by  cutting  from  each  corner  a 
square  of  side  b  inches.     Find  the  dimensions  of  the  original  piece  of  tin. 

12.  One  edge  of  a  rectangular  box  is  increased  6  in.,  another  3  in.,  and 
the  third  4  in.,  making  a  cube  whose  volume  is  862  cu.  in.  greater  than 
that  of  the  original  box.     Find  the  dimensions  of  the  box. 

Arts.     4.14  in.  by  7.14  in.  by  6.14  in. 

13.  If  in  a  right  prism  the  altitude  is  equal  to  a  side  of  the  base,  find 
the  volume  if  the  base  is  an  equilateral  triangle  whose  sides  are  a. 

14.  Find  the  capacity  in  gallons  of  a  water    _^~~7 — 
tank  for  a  locomotive  tender.     The  dimensions  J 
are  as  given  in  the  figure  for  the  length  cross 
section,  and  the  width  is  9  ft.  6  in. 

15.  A  certain  coal  and  coke  company  finds  it  necessary 
to  construct  a  wharf  wall  150  ft.  long,  and  having  a  cross 
section  with  dimensions  as  shown  in  the  figure.  Find  the 
number  of  cubic  yards  in  the  wall. 

By  "  batter  1:12"  is  meant  that  the  slant  is  1  ft.  in  a 
vertical  rise  of  12  ft. 

16.  In  a  right  prism  whose  volume  is  54,  the  lateral  area 
is  four  times  the  area  of  the  base  which  is  an  equilateral 
triangle.     Find  the  edge  of  the  base.  Ans.     6. 

17.  The  total  areas  of  two  similar  cylinders  of  revolu- 
tion are  75  sq.  in.  and  192  sq.  in.  respectively.     If  the  volume  of  the  first 
cylinder  is  250  cu.  in.,  what  is  the  volume  of  the  second? 

18.  In  a  right  circular  cylinder,  given  V —TTr2h  and  T  =  2lTr2-{-2Trh. 

Find  T  in  terms  of  V  and  h.  _     27 

Ans. 

19.  In  a  right  circular  cylinder,  find  V  in  terms  of  the  circumference 


c  of  the  base  and  the  total  area  T. 


Ans.     V=- 


2wcT-c3 


87T2 


20.  Find  the  steam  capacity  of  a  horizontal  cylindrical  boiler  4  ft.  in 
diameter  and  16  ft.  long,  if  the  height  of  the  segment  occupied  by  the 
steam  is  18  in.      (Use  (2)  of  §509). 


GENERAL  EXERCISES 


341 


21.  A  cylindrical  tank  of  diameter  30  in.  and  34  in.  long  rests  on  its 
side.  Find  the  number  of  gallons  of  gasoline  in  the  tank  if  the  depth  is 
5j  in.  Ans.     13  gal.  nearly. 

22.  The  volume  of  an  irregular  shaped  body  is  often 
found  by  immersing  it  in  water  and  determining  the 
amount  of  water  displaced.  A  cylindrical  vessel  that  has 
a  diameter  of  4  in.  is  partly  filled  with  water.  A  stone 
immersed  in  the  water  raises  its  level  3 \  in.  Find  the  vol- 
ume of  the  stone. 

23.  A  silo  is  used  to  keep  fodder  in  a  green  and  succulent  state  for  feed- 
ing farm  animals.     It  is  usually  built  in  a  cylindrical  form. 

Find  the  capacity  in  tons  of  a  silo  in  the  form  of  a  right  circular  cylinder 
20  ft.  in  diameter  and  32  ft.  high,  if  a  cubic  foot  of  silage  weighs  40.7  lb. 

24.  An  18  acre  field  yields  11.5  tons  of  silage  per  acre.  What  must  be 
the  height  of  a  cylindrical  silo  20  ft.  in  diameter  to  hold  all  the  silage  if 
1  cu.  ft.  of  silage  weighs  38.4  lb.? 

25.  A  silo  is  in  the  form  of  a  right  circular  cylinder,  and  is  20  ft.  in 
diameter  inside  and  32  ft.  high.     How  many  cubic  yards  of  concrete  did 

'  it  take  to  build  it,  if  the  floor  and  wall  are  each  6  in.  thick,  and  the  founda- 
I  tion  wall  is  8  in.  thick  and  5  ft.  deep?     ^ 18'1 >h-sK— 

26.  The  connecting  rod  with  dimen-  P ■ —  \\» 

sions  as  given  in  the  figure  is  made  from  Lr 

stock  with  the  dimensions  shown.     Find 

length   of   stock    that   it   is   necessary  Connecting  Rod 

to  allow  for  the  cylindrical  part  of  the 

rod.  stock 

27.  Find  the  volume  of  the  beveled  washer  with  dimensions  as  shown 
in  the  figure. 

28.  A  pine  log  2  ft.  in  diameter  and  16  ft. 
long  is  floating  in  water.  Find  the  weight  of 
the  log  if  two-thirds  of  the  volume  of  the  log  is 
under  water.  (Water  weighs  62.5  lb.  per 
cubic  foot.) 

29.  Will  a  floating  pine  log  lj  ft.  in  diam- 
eter and  10  ft.  long  support  a  man  weighing  180  pounds  if  the  specific 
gravity  of  the  log  is  0.72? 

30.  The  following  is  the  record  of  a  test  in  which  a  high  speed  drill 
removed  70.56  cu.  in.  of  cast-iron  per  minute.  The  penetration  per  minute 
was  57 J  in.,  the  feed  y^-  in.  per  turn,  and  the  number  of  turns  per 
minute  was  575.     Do  these  numbers  agree? 


(25" 


5 

LU 


342  SOLID  GEOMETRY 

THEOREMS  AND  PROBLEMS 

1.  Derive  the  following  formulas  for  finding  th3  volume,  V,  of  a  hollow 
circular  cylinder  of  length  h  and  cross  sectional  dimensions  as  given  in  the; 
figure: 

(1)  V  =  7Th(R+r)(R-r). 

(2)  V  =  lTTht(D+d). 

(3)  V  =  Tht(d+t). 

(4)  V  =  7Tht(D-t). 

2.  Construct  a  plane  through  a  point  and  tangent  to 
a  given  right  circular  cylinder. 


3.  If  the  radius  of  one  cylinder  is  equal  to  the  alti- 
tude  of  a  second,  and  the  radius  of  the  second  is  equal  to  the  altitude  of 
the  first,  what  is  the  ratio  of  their  volumes? 

4.  The  intersection  of  two  planes  each  tangent  to  a  circular  cylinder  is 
parallel  to  the  elements  of  the  cylinder. 

5.  Prove  that  the  volume  of  a  right  circular  cylinder  is  equal  to  its 
lateral  area  times  one-half  the  radius  of  its  base. 

6.  The  volume  of  two  right  circular  cylinders  are  equal.     Write  a 
proportion  between  their  lateral  areas  and  their  radii. 

7.  If  a  straight  line  has  morethan  two  points  common  to  the  curved 
surface  of  a  right  circular  cylinder,  the  fine  is  an  element  of  the  surface. 


CHAPTER  VIII 
PYRAMIDS  AND   CONES 


693.  A  moving  straight  line  that  always  contains  a  fixed 
point,  and  always  intersects  a  given  straight  line,  generates  a 
plane.     Why?  §542 

694.  Pyramidal  Surface.  A  moving  straight  line  that 
always  contains  a  fixed  point,  and  always  intersects  a  broken 
line  not  coplanar  with  it,  generates  a  pyramidal  surface. 


695.  Conical  Surface.  A  moving  straight  line  that  always 
contains  a  fixed  point,  and  always  intersects  a  plane  curved 
line  not  coplanar  with  it,  generates  a  conical  surface. 

696.  The  fixed  point  is  called  the  vertex  of  the  pyramidal, 
or  conical,  surface. 

697.  The  two  parts  of  the  pyramidal,  or  conical,  surface  on 
opposite  sides  of  the  vertex  are  called  nappes. 

The  words  generatrix,  directrix,  element,  and  closed  surface  have  the 
same  significance  here  as  in  §§618,  619. 

The  directrix  is  not  necessarily  closed,  but  in  this  text  only  those  that 
are  closed  are  considered. 

The  word  " section"  has  the  same  significance  as  before  (§  620),  how- 
ever, the  cutting  plane  must  not  pass  through  the  vertex. 

343       ' 


344 


SOLID  GEOMETRY 


698.  A  pyramid  is  the  solid  formed  by  cutting  all  the  ele- 
ments  of   one    nappe    of    a 
closed  pyramidal  surface  by 
a  plane. 

699.  A  cone  is  the  solid 
form  by  cutting  all  the  ele- 
ments of  one  nappe  of  a 
closed   conical   surface  by  a  plane. 

The  meanings  of  the  words:  base,  lateral  face,  lateral  edge,  base  edge, 
lateral  area,  total  area,  are  apparent  from  the  definitions  of  §§  623,  624. 

700.  The  altitude  of  a  pyramid,  or  cone,  is  the  perpendic- 
ular distance  from  the  vertex  to  the  base. 

701.  Pyramids  Classified  According  to  the  Number  of  Lat- 
eral Faces.  Pyramids,  like  prisms,  are  classified  as  triangular, 
quadrangular,  pentagonal,  etc.,  according  as  their  bases  are 
triangles,  quadrilaterals,  pentagons,  etc. 

702.  Regular  Pyramids.    A  pyramid  whose  base  is  a  regu 
lar   polygon,    and   whose   altitude   meets    the 
center  of  its  base,  is  called  a  regular  pyramid. 

The  altitude  of  a  regular  pyramid  is  called 
its  axis. 

The  altitude  of  one  of  the  lateral  faces  of  a 
regular  pyramid  is  called  the  slant  height. 

In  the  figure,  OF  is  the  altitude,  BV  a  lateral  edge,        B  C 

and  NV  the  slant  height.  Regular  Pyramii 

703.  Circular  Cones.     A  cone  whose  base  is  a  circle 
called  a  circular  cone. 

The  line  joining  the  vertex 
of  a  circular  cone  to  the  center 
of  its  base  is  called  the  axis  of 
the  cone. 

704.  A  right  circular  cone  is 
a  circular  cone  whose  axis  is 

,.      ,  .,     ,  Circular  Right-Circular 

perpendicular  to  its  base.  Cone  Cone 


PYRAMIDS  AND  CONES 


345 


The  radius  of  a  circular  cone  is  the  radius  of  its  base. 

Since  a  right  circular  cone  may  be  generated  by 
revolving  a  right  triangle  about  one  of  its  sides  as 
an  axis  it  is  often  called  a  cone  of  revolution. 

The  length  of  an  element  of  a  right  circular 
cone  is  its  slant  height. 

705.    Prove  the  following  facts  concerning  cones  , 
and  pyramids: 

(1)  The  lateral  faces  of  a  regular  pyramid  are  congruent  isos- 
celes triangles. 

(2)  The  lateral  edges  of  a  regular  pyramid  are  equal. 

(3)  The  elements  of  a  right  circular  cone  are  equal. 

(4)  The  axis  of  a  right  circular  cone  coincides  with  its  altitude. 

(5)  A  straight  line  drawn  from  the  vertex  of  a  cone  to  any 
point  in  the  perimeter  of  its  base  is  an  element. 

(6)  The  section  of  a  circular  cone  made  by  a  plane  containing 
an  element  is  a  triangle. 

(7)  The  section  of  a  pyramid  made  by  a  plane  through  its 
vertex  is  a  triangle. 

EXERCISES 

1.  Find  an  element  of  a  right  circular  cone  whose  altitude  is  15  and 
radius  7. 

2.  Find  the  altitude  of  a  right  circular  cone  whose 
slant  height  is  s  and  radius  r. 

3.  Find  the  altitude  of  a  regular  pyramid  each 
face  and  the  base  being  an  equilateral  triangle  10  in. 
on  a  side. 

4.  Into  how  many  parts  do  the  nappes  of  a  conical 
surface  divide  space? 

5.  Why  is  it  stated  in  the  definition  of  a  pyramidal 
surface  that  the  vertex  and  directrix  must  not  be 
coplanar? 

6.  The  altitude  of  a  right  circular  cone  is  10  in. 
and  the  radius  of  the  base  is  6  in.  Find  the  area  of  a 
section  made  by  a  plane  passing  through  the  vertex 
and  3  in.  from  the  center  of  the  base.  In  the  figure 
find  the  area  of  VAB. 


346 


SOLID  GEOMETRY 


706.  Theorem.  The  lateral  edges  and  the  altitude  of 
mid,  or  the  elements  and  the  altitude  of  a  cone,  are  divided 
tionally  by  a  plane  parallel  to  its  base. 


a  pyra 
propor- 


Pass  a  plane  through  the  vertex  parallel  to  the  base  and  apply  §  568. 

707.    Theorem.     The  section  of  a  pyramid  made  by  a  plane 
parallel  to  the  base  is  similar  to  the  base. 

Given  pyramid   V-AD,   cut  by  plane   P 
parallel  to  base  AD,  forming  the  section  FI. 

To  prove  polygon  Fl^polygon  AD. 

Proof.     AFVG^AAVB,  AGVH^ABVC, 
etc.  Why? 


^      FG     VG        .  GH     VG 

Then = ,  and  —  = ,  etc. 

AB     VB  BC     VB 


Hence 


Also 


§428 
JF 


FG_GH_HI_IJ  _ 
AB~  BC    CD    DE    EA 
ZJFG  =  ZEAB,  ZFGH  =  ZABC,  etc. 
.'.  polygon  W^polygon  AD. 


Why? 

§567 
§§  441,  43| 


EXERCISES 

1.  The  lateral  edges  of  a  pyramid  are  respectively  15  in.,  12  in.,  and 
16  in.  Find  the  parts  of  each  made  by  a  plane  that  is  parallel  to  the  base 
and  which  divides  the  altitude  into  parts  that  are  in  the  ratio  of  1  :  [I 

2.  The  area  of  the  base  of  a  pyramid  is  125  sq.  in.  Find  the  area  of 
a  section  8  in.  from  the  base  and  parallel  to  it,  if  two  corresponding  edges 
of  the  base  and  the  section  are  respectively  10  in.  and  8  in. 


PYRAMIDS  AND  CONES 


347 


708.  Theorem.  The  section  of  a  circular  cone  made  by  a  "plane 
parallel  to  its  base  is  a  circle,  the  center  of  which  is  the  point  where 
the  axis  intersects  it. 

Given  a  circular  cone  V-ABC,  with  section 
EH  made  by  the  plane  P  parallel  to  the  base. 
Also  given  axis  VO  piercing  plane  P  in  0'. 

To  prove  that  EH  is  a  circle  having  0'  as 
center. 

Proof.  Take  F  and  G  any  two  points  in 
EH.  M 

Then  the  planes  determined  by  FO'V  B 
and  GO'V  intersect  the  plane  of  the  base  in  OB  and  OC,  the 
plane  P  in  O'F  and  O'G,  and  the  lateral  surface  in  the  elements 
VB  and  VC,  respectively.  Why? 

Prove  that  AFO'V^ABOV,  and  AGO'V~ ACOV. 
FO'_GO' 
BO~ CO' 

But  BO  =  CO.  Why? 

Therefore  FO'  =  GO'  Why? 

Since  F  and  G  are  any  two  points  in  the  section  EH,  all  points 
in  this  section  are  equally  distant  from  0' . 

.*.  section  EH  is  a  circle  with  center  0' '.  §  263 

709.  Theorem.  The  area  of  any  section  of  a  pyramid,  or 
circular  cone,  parallel  to  the  base  is  to  the  area  of  the  base  as  the 
square  of  its  distance  from  the  vertex  is  to  the  square  of  the  altitude 
of  the  pyramid,  or  cone. 


Then 


Why? 


v 


EXERCISES 

1.  The  area  of  the  base  of  a  pyramid  is  64  sq.  in.  and  its  altitude  is 
20  in.  Find  the  area  of  a  section  parallel  to  the  base  and  4  in.  from  the 
vertex. 

2.  The  diameter  of  the  base  of  a  circular  cone  is  6  in.  and  the  altitude 
is  10  in.  Find  the  area  of  a  section  parallel  to  the  base  and  6  in.  from  the 
vertex. 


348 


SOLID   GEOMETRY 


710.  Theorem.  2/  two  pyramids,  two  circular  cones,  or  a 
pyramid  and  a  circular  cone,  have  equal  altitudes  and  bases  of 
equal  area,  sections  made  by  planes  parallel  to  the  bases  and  at 
equal  distances  from  the  vertices,  have  equal  areas. 


Suggestion.     Let  the  equal  altitudes  be  h  and  the  distance 
from  the  vertices  to  the  sections  h! . 

Then  area  of  any  section  is  to  area  of  base  as  h'2  :  h2.  §  709 
But  it  is  given  that  the  bases  are  equal,  and  the  altitudes  equal. 
Therefore  the  sections  are  equal  in  area.  Why? 


EXERCISES 

1.  In  the  figure  of  §707,  VO'  =  8,  0'0  =  12.     F£  =  25,  and  VA  = 
find  VG  and  VF. 

2.  If  the  base  of  the  pyramid  of  §  707  has  an  area  of  180  sq.  in.,  using 
the  lengths  given  in  Exercise  1,  find  the  area  of  the  section. 

3.  The  area  of  the  base  of  a  pyramid  is  150  sq.  cm.  and  its  altitude  is 
2  dm.     Find  the  area  of  a  section  8  cm.  from  the  vertex. 

4.  The  areas  of  parallel  sections  of  a  pyramid,  or  of  a  circular  cone, 
are  to  each  other  as  the  squares  of  their  distances  from  the  vertex. 

5.  The  altitude  of  a  pyramid  is  h.  At  what  distance  from  the  vertex 
must  a  plane  be  passed  parallel  to  the  base  so  that  the  section  shall  be  (1) 
one-half  as  large  as  the  base?     (2)  One-third?     (3)  One-fifth? 

6.  Each  side  of  the  base  of  a  regular  hexagonal  pyramid  is  8  in. 
the  altitude  of  the  pyramid  is  20  in.,  how  far  from  the  vertex  must  a  plane 
parallel  to  the  base  be  so  that  the  area  of  the  section  shall  be  24 -\/3  sq.  in.? 

7.  The  area  of  the  base  of  a  cone  is  216  sq.  in.  and  its  altitude  is  24  in. 
Find  the  distance  between  two  sections  parallel  to  the  base,  which  have 
areas  of  144  sq.  in.  and  72  sq.  in.,  respectively. 


If 


AREAS  OF  PYRAMIDS  AND  CONES 


349 


AREAS   OF   PYRAMIDS   AND   CONES 

711.  Theorem.  The  lateral  area  of  a  regular  pyramid  is 
equal  to  half  the  product  of  the  perimeter  of  the  base  and  the 
slant  height.    S  =  \ps. 

Given  the  regular  pyramid  V-ABCDE. 

To  prove  S  =  ^ps,  where  S  denotes  lateral 
area,  s  slant  height,  and  p  perimeter  of  base. 

Proof.     The  lateral  faces  are  congruent  tri- 
angles. Why? 

The  area  of  each  face  triangle  =  |sX its  base. 

Why? 

The  sum  of  the  bases  of  the  triangles  =  p. 

Hence  the  sum  of  the  areas  of  the  faces  =  ^ps 
.'.S  =  %ps. 


Why? 
Why? 


EXERCISES 

as  2S 

1.  Using  the  notation  of  §711  show  that  s= — ,  and  p= — . 

p  s 

2.  Find  the  lateral  area  and  the  total  area  of  a  regular  hexagonal  pyra- 
mid whose  altitude  is  8  in.  and  base  edge  is  4\/3  in. 

3.  Find  the  lateral  area  and  the  total  area  of  a  regular  triangular 
pyramid  having  base  edges  equal  to  8  in.  and  lateral  edges  equal  to  12  in. 

712.  Plane  Tangent  to  a  Cone.     A  plane  is  tangent  to  a 
cone  if  it  meets  the  conical  surface  in  one  ele- 
ment only. 

713.  Theorem.  If  a  plane  is  tangent  to  a 
circular  cone,  its  intersection  with  the  plane  of 
the  base  is  tangent  to  the  base. 

Proof  similar  to  that  of  §  642. 

714.  Theorem.  The  plane  determined  by 
the  tangent  to  the  base  of  a  circular  cone  and  the 
element  through  the  point  of  contact,  is  tangent  to  the  cone. 

Compare  with  theorem  of  §  643. 


350 


SOLID   GEOMETRY 


715.  Inscribed  and  Circumscribed  Pyramids.  A  pyramid 
is  said  to  be  inscribed  in  a  cone  if  the  pyramid  and  cone  have 
the  same  vertex,  and  if  the  base  of  the  pyramid  is  inscribed 
in  the  base  of  the  cone.  The  cone  is  then  said  to  be  circum- 
scribed about  the  pyramid. 


716.  A  pyramid  is  said  to  be  circumscribed  about  a  cone 
if  the  pyramid  and  cone  have  the  same  vertex,  and  if  the 
base  of  the  pyramid  is  circumscribed  about  the  base  of  the 
cone.    The  cone  is  then  said  to  be  inscribed  in  the  pyramid. 

It  is  evident  that  the  lateral  edges  of  an  inscribed  pyramid  are  elements 
of  the  cone;  and  that  the  lateral  faces  of  the  circumscribed  pyramid  are 
tangent  to  the  cone  by  §§712,  715. 

717.  Area  of  Cone,  Practical  Consideration.  If  the  lateral 
surface  of  a  right  circular  cone  is  covered  with  paper,  this  paper 
when  peeled  from  the  cone  can  be  spread  out  and  will  be  a  sec- 
tor of  a  circle,  whose  radius  is  the  slant  height  of  the  cone,  and 


whose  arc  is  equal  to  the  circumference  of  the  base  of  the  cone. 
Then,  by  §  508,  the  lateral  area  of  a  right  circular  cone  is  equal  to 
half  the  product  of  the  slant  height  and  the  circumference  of  the 


base.     S  =  ^sc=irrs. 


AREAS  OF  PYRAMIDS  AND  CONES  351 

718.  Geometric  Treatment.  A  method  for  measuring  the 
surface  of  a  cone,  which  is  accurate  and  geometric  follows 
from  a  consideration  similar  to  that  of  §  649.  In  this  manner 
the  truth  of  the  following  statement  is  established. 

The  lateral  area  of  a  right  circular  cone  is  the  common  limit  of 
the  areas  of  inscribed  and  circumscribed  regular  pyramids,  as  the 
number  of  faces  is  indefinitely  doubled. 

719.  Theorem.  The  lateral  area  of  a  right  circular  cone  is 
equal  to  half  the  product  of  the  circumference  of  the  base  and 
the  slant  height.     S  =  ^cs=7rrs.  A 

Given  a  right  circular  cone. 

To  prove  S  =  ^cs= wrs,  where  S  denotes  lateral  J 

area,  s  slant  height,  c  circumference  of  base,        Mi 
and  r  radius.  IMl 

Proof.  Circumscribe  a  regular  pyramid  MmA^^^^\ 
about  the  cone,  and  let  S'  denote  its  lateral  vfiw/  luuOT^ 
area  and  p  the  perimeter  of  its  base. 

By  doubling  indefinitely  the  number  of  faces  of  the  pyramid, 
S'->S,  and  p->c.  §§  718,  488 

Then  \ps-*\cs.  §  485  (2) 

But      S'=^ps,  being  variables  that  are  always  equal.    §  711 
,\S=4<».  §485(1) 

Also  c  =  27rr.  §495 

.\  S=Trrs.  '       §111 

Remark.  It  is  to  be  noted  that  the  formulas  of  this  article 
apply  only  to  right  circular  cones. 

720.  Theorem.  The  total  area  T  of  a  right  circular  cone  is 
given  by  the  formula  T =Trrs+irr2 =7rr(s+r) . 

EXERCISES 

1.  Form  a  cone  from  a  semicircle  having  a  radius  of  3  in.  Find  the 
lateral  area  of  the  cone.     The  total  area. 

2.  Show  how  to  cut  a  pattern  for  a  tin  cone  that  is  to  have  the  diame- 
ter of  its  base  4  in.  and  its  slant  height  10  in. 


352  SOLID  GEOMETRY 

2S     S         2S        %        8 

3.  In  a  right  circular  cone  show  that  s  = — = — ,  c— — ,  and  r= — . 

c      7Tr  s  7Ts 

4.  Solve    T  =  Wrs+Trr2     and     find     r  = ,      and 

T-7Tr\    T  27r 

7Tr         7Tr 
In  a  right  circular  cone,  using  S  for  lateral  area,  T  for  total  area,  s  for 
slant  height,  r  for  radius,  c  for  circumference,  and  h  for  altitude,  solve  the 
following : 

5.  Given  r  =  4  in.,  s  =  6  in.;  find  S  and  T. 

6.  Given  c  =  25  in.,  s  =  8  in.;  find  S  and  T. 

7.  Given  s  =  8  ft.  6  in.,  c  =  3  ft.  7  in. ;  find  S. 

8.  Given  s  =  7.2  in.,  r  =  5.3  in.;  find  S  and  T\ 

9.  Given  h  =  12  in.,  r  =  5  in.;  find  S  and  T. 

10.  Given  c  =  30  in.,  s  =  8  in.;  find  h  and  &. 

11.  Given  £  =  200  sq.  in.,  r  =  6  in.;  find  s. 

12.  Given  £  =  256  sq.  cm.,  s  =  16  cm.;  find  r. 

13.  Given  T  =  500  sq.  cm.,  r  =  10  cm.;  find  s. 

14.  The  circumference  of  the  base  of  a  conical  church  steeple  is  35  ft. 
and  the  altitude  is  73  ft.     Find  the  lateral  area.  Ans.     1281— sq.ft. 

15.  Find  the  lateral  edge  and  the  lateral  area  of  a  regular  pyramid 
each  side  of  whose  triangular  base  is  10  ft.,  and  whose  altitude  is  18  ft. 

Ans.     18.90+  ft.,  273.45-  sq.  ft. 

16.  A  tower  with  a  regular  hexagonal  base  has  dimensions 
as  shown  in  the  figure.  The  pitch  of  the  pyramdial  roof  is  1  \> 
which  means  that  OV  =  1^  QR.  Find  the  lateral  area  of  the 
tower. 

17.  Cut  out  of  heavy  paper  a  sector  of  a  circle,  with  a  radius 
of  3  in.  and  the  central  angle  120°.  Bring  the  edges  together 
and  paste  them.  Find  the  lateral  area  and  the  total  area  of 
the  cone  thus  formed. 

18.  The  cone  formed  by  revolving  an  equilateral  triangle 
about  one  of  its  altitudes  has  a  lateral  area  equal  to  twice  the  U ' 
area  of  the  base. 

19.  Find  the  area  of  the  surface  of  the  solid  formed  by  revolving  an 
equilateral  triangle,  having  a  side  of  12  in.,  about  one  of  its  sides. 

20.  Find  the  area  of  the  surface  of  the  solid  formed  by  revolving  a 
right  triangle,  having  a  base  of  8  in.  and  an  altitude  of  6  in.,  about  its 
hypotenuse. 


AREA  OF   FRUSTUM   OF  PYRAMID  OR  CONE         353 

AREA   OF  FRUSTUM   OF   PYRAMID    OR   CONE 

721.  Truncated  Pyramid,  or  Cone.  The  portion  of  a  pyra- 
mid, or  cone,  included  between  the  base  and  a  section  not 
parallel  to  the  base  is  called  a  truncated  pyramid,  or  cone. 

722.  Frustum.  The  portion  of  a  pyramid,  or  cone,  included 
between  the  base  and  a  section  parallel  to  the  base  is  called  a 
frustum  of  the  pyramid,  or  cone. 


723.  The  base  and  the  parallel  section  are  called  the  bases 
of  the  frustum.  The  perpendicular  distance  between  the  bases 
is  the  altitude  of  the  frustum. 

724.  The  slant  height  of  a  frustum  of  a  regular  pyramid, 
or  of  a  cone,  is  that  portion  of  the  slant  height  of  the  pyra- 
mid, or  cone,  included  between  the  bases  of  the  frustum. 

Thus  AB  and  CD  are  slant  heights,  and  QP  and  MN  are  altitudes. 

725.  Midsection.  If  a  solid  has  parallel  bases,  the  section 
parallel  to  the  bases  and  half  way  between  them  is  called  the 
midsection.  a     ~ — ^f 

In  the  figure,   NR  is  the  midsection. 

726.  Prove  the  following  facts  concern- 
ing frustums: 

(1)  .The  lateral  faces  of  a  frustum  of  a  D^ 
regular    pyramid    are    congruent    isosceles 
trapezoids. 

(2)  The  lateral  edges  of  a  frustum  of  a  regular  pyramid  are 
equal;  and  the  slant  height  is  the  same  for  all  the  faces. 


S54 


SOLID  GEOMETRY 


727.  Theorem.  The  lateral  area  of  a  frustum  of  a  regular 
pyramid  is  equal  to  half  the  product  of  the  slant  height  and  the 
sum  of  the  perimeters  of  the  bases.    S  =  ^s(P+p). 

Given  the  frustum  of  a  regular  pyramid 
with  bases  AC  and  EG. 

To  prove  that  S  =  %s(P+p),  where  8 
denotes  lateral  area,  s  slant  height,  and  P 
and  p  the  perimeter  of  the  lower  and  upper 
base  respectively. 


Proof. 


Hence 
But 


Area  AF=±s(AB+EF). 
Avesi  BG  =  ^s(BC+FG). 
etc. 

S  =  %s[(AB+BC+---)  +  (EF+FG+- 
AB+BC+  •  •  •  =  P,  and  EF+FG+  •  •  • 

.•.s=4*(p+p). 


=p. 


728.  Theorem.  The  lateral  area  of  a  frustum  of 
circular  cone  is  equal  to  half  the  product  of  the  slant 
and  the  sum  of  the  circumferences  of  the  bases. 

Given  the  frustum  of  a  right  circular  cone. 

To  prove  S=ir(R+r)s,  where  S  denotes 
lateral  area,  s  slant  height,  and  R  and  r  the 
radius  of  the  lower  and  upper  base  respec- 
tively. 

Proof.  Let  m  be  the  slant  height  of  the 
cone  and  n  the  slant  height  of  the  part  above 
the  frustum. 

Then  >S  =  mirR — rnrr = ir  (mR  —  nr) 

But  R  :  r  =  m  :  n, 

And  therefore  mr  —  nR  =  0 . 

Hence  S=ir(mR  —  nr-\-mr  —  nR) 

—ir{R-\-r)  (?n  —  7i). 

But  m  —  n  =  s. 

.\S**ir{R+ty$. 


§370 


§105 
Why? 
Why? 

a  right 
height 


Why? 

§428 
§398 

Why? 

Why? 

Why? 


AREA  OF  FRUSTUM  OF  PYRAMID  , OR  GC^T?' ;,  | ;  J^&x  i 

729.  Theorem.  The  lateral  area  of  a  frustum  of  a  right  cir- 
cular cone  is  equal  to  the  product  of  the  altitude  and  the  circumfer- 
ence of  a  circle  whose  radius  is  the  perpendicular  at  the  midpoint 
of  an  element,  and  terminated  by  the  axis.  ^  r 

Outline  of  proof.     Let  a  be  the  length  of  EF,         \\v\\ 
the  perpendicular  at  the  midpoint  of  an  element     /  hD\  r>JU 
and  terminated  by  the  axis.     Let  r'  be  the  /      Er±  \\ 

radius  of  the  midsection,  and  s  and  h  the  slant  AllLi ClA->* 

height  and  altitude  respectively. 

Show  that  ADEF^AABC. 

Then  s  :  h  =  a  :  r',  and  srf  =  ah. 

Show  that  R+r  =  2r',  where  R  and  r  are  the  radii  of  the  bases. 

Then,  substituting  in  the  formula  of  §  728,  S  =  2irah. 

EXERCISES 

1.  Prove  the  theorem  of  §  728  by  considering  the  lateral  area  of  a 
frustum  of  a  right  circular  cone  as  the  limit  of  the  area  of  a  frustum  of 

pyramid. 

2.  Prove  the  theorem  of  §  727  by  using  a  method  similar  to  that  of 
§  728  for  the  frustum  of  a  cone. 

3.  Show  that  the  lateral  area  of  a  frustum  of  a  right  circular  cone  is 
equal  to  the  slant  height  times  the  perimeter  of  the  midsection. 

4.  Find  the  lateral  area  of  the  frustum  of  a  regular  quadrangular 
pyramid,  if  an  edge  of  the  lower  base  is  16  in.,  an  edge  of  the  upper  base 
12  in.,  and  the  slant  height  18  in. 

5.  Find  the  total  area  of  the  frustum  of  a  right  circular  cone,  if  the 
radii  of  the  bases  are  6  ft.  and  8  ft.  respectively,  and  the  slant  height  is  7  ft. 

6.  A  portion  of  the  roof  of  a  tower  is  a  frustum  of  a  right  circular 
cone.  The  radii  of  the  bases  are  10  ft.  and  6  ft.  respectively,  and  the  alti- 
tude is  8  ft.     Find  the  number  of  square  feet  in  this  part  of  the  roof. 

7.  The  altitude  of  a  right  circular  cone  is  h.  How  far  from  the  vertex 
must  a  plane  be  passed  parallel  to  the  base  so  that  the  lateral  area  of  the 
cone  cut  off  shall  equal  the  lateral  area  of  the  frustum.     Arts.     \hy/2. 

8.  The  diameter  of  the  bottom  of  a  pail  is  10  in.  and  that  of  the  top 
12  in.  Find  the  number  of  square  inches  of  tin  in  the  pail  if  the  slant 
height  is  11  in.  Arts.     458.67. 


:  856 1 


-.SOLID   GEOMETRY 


9.  Determine  the  diameter  of  the  blank  to  make  a 
pressed  basin  of  the  form  shown  in  the  figure.  The  depth 
is  3  in.,  the  bottom  has  a  diameter  of  6  in.,  the  top  an 
inside  diameter  of  7  in.,  and  the  rim  is  J  in.  wide. 

Ans.     11.4+  in. 
Suggestion.     Consider  the  rim  as  a  ring  between  two  concentric  circles. 
The  blank  must  have  an  area  equal  to  the  total  area  of  the  basin. 

10.  A  pie  tin  has  a  diameter  of  7 \  in.  at  the  bottom  and  9  in.  at  the  top 
inside.  It  is  1  in.  deep  and  has  a  flange  J-  in.  wide  around  the  edge.  Find 
the  diameter  of  the  blank  required  to  make  the  tin. 

11.  Find  the  area  generated  by  revolving  a  square  a  inches  on  a  side 
about  a  diagonal. 

12.  A  tower  whose  cross  section  is  a  regular  octagon  6  ft.  on  a  side  has 
a  roof  that  is  full  pitch.     Find  the  area  of  the  roof. 

By  full  pitch  is  meant  that  the  vertex  of  the  roof  is  the  same  distance 
above  the  plates  as  the  width  of  the  tower  from  face  to  face. 

13.  The  conical  roof  over  a  water  tank  is  half  pitch.  Find  its  area  if 
the  diameter  of  the  tank  is  12  ft.  and  the  roof  projects  1^  ft.  at  the  eaves. 

Ans.     221.5- sq.ft. 
By  half  pitch  is  meant  that  the  vertex  of  the  roof  is  one-half  the  diameter 
of  the  tower  above  the  top  of  the  tower. 

14.  Find  the  total  area  of  the  solid  generated  by  revolving  an  isosceles 
trapezoid  about  a  line  connecting  the  middle  points  of  its  bases,  if  the 
bases  are  8  in.  and  12  in.  respectively,  and  the  angles  at  one  base  are  each 
60°. 

15.  Determine  how  to  draw  a    IV 

c  C 
pattern  for  the  upper  part  of  the     v 

funnel  with  dimensions  as  shown, 

the  diameter  at  A  being  J  in.  and  at 

B,   1    in.     The    allowance   for    the 

seam  can  be  added. 

Suggestion.  The  length  OP  =  x 
may  be  determined  as  follows: 
x  :z+3  =  l  :  4.      .-.  x  =  l. 

The  radius  to  use  is  therefore  4  in. 


To  determine  the  central  angle  u, 


QnR  _    u 
~87r~~360° 


But  QnR  =  47T.      .\u  =  180°. 

16.    Determine  how  to  draw  a  pattern  for  the  lower  part  of  the  funnel 
shown  in  the  figure  of  the  previous  exercise. 


VOLUMES  OF  PYRAMIDS  ANfr  CONES 
VOLUMES   OF  PYRAMIDS   AND    CONES 


3o7: 


730.  Inscribed  Prisms.  If  the  altitude  of  a  triangular 
pyramid  is  divided  into  equal  parts  by  a  series  of  planes  par- 
allel to  the  base,  the  prisms  within  the  pyramid  and  having 
the  sections  formed  by  the  parallel  planes  as  bases,  and  each 
having  one  of  the  segments  of  one  of  the  lateral  edges  of 
the  pyramid  as  one  of  its  lateral  edges,  are  called  a  set  of 
inscribed  prisms. 

In  the  figure  M ,  N,  and  R  are  a  set  of  inscribed  prisms,  each 
being  entirely  within  the  pyramid. 


731.  Circumscribed  Prisms.  In  a  similar  manner,  the 
prisms  having  the  base  of  the  pyramid  and  the  sections  as 
bases  are  called  a  set  of  circumscribed  prisms. 

In  the  figure,  S,  T,  U,  and  V  are  a  set  of  circumscribed 
prisms,  each  being  partly  outside  of  the  pyramid. 

732.  The  two  sets  of  such  prisms,  formed  by  using  the  same 
lateral  edge  and  the  same  parallel  planes,  are  called  corre- 
sponding sets  of  inscribed  and  circumscribed  prisms. 

733.  The  following  statement  may  be  considered  as  evident 
from  the  preceding: 

The  volume  of  a  triangular  pyramid  is  definite,  and  is  greater 
than  that  of  any  set  of  inscribed  prisms,  and  less  than  that  of  any 
set  of  circumscribed  prisms. 


;  ggg ! ;  :  / '  •, '  ! ;  \      :  $OLID  GEOMETRY 

734.  Theorem.  The  difference  in  volume  between  a  set  of 
inscribed  prisms  and  the  corresponding  set  of  circumscribed  prisms 
is  the  volume  of  the  prism  upon  the  base  of  the  pyramid. 

735.  Theorem.  The  volume  of  a  triangular  pyramid  is  the 
common  limit  of  the  volumes  of  the  set  of  inscribed  prisms  and  the 
corresponding  set  of  circumscribed  prisms,  as  the  number  of  these 
prisms  is  increased  indefinitely. 

Given  the  triangular  pyramid  D-ABC,  a  set  of  inscribed 
prisms,  and  corresponding  set  of  circumscribed  prisms. 

To  prove  that  volume  of  D-ABC  is  the  common  limit  of  these 
two  corresponding  sets  of  prisms.  t^p^d 

Outline  of  proof.  The  volume  of  the 
triangular  pyramid  is  greater  than  the  sum 
of  the  inscribed  prisms  and  less  than  the 
sum  of  the  circumscribed  prisms.         §  733 

Therefore,  the  volume  of  the  pyramid 
differs  from  either  by  an  amount  less  than 
the  volume  of  the  prism  upon  the  base  of 
the  pyramid.  §  Why? 

But  the  volume  of  this  prism  can  be  made  as  small  as  desired, 
that  is,  it  has  zero  for  a  limit.  §  485  (2) 

Hence,  the  volume  of  the  set  of  inscribed  prisms  and  the  set 
of  circumscribed  prisms,  each  have  the  volume  of  the  triangular 
pyramid  for  a  limit.  §  483 

EXERCISES 

1.  Corresponding  sets  of  inscribed  and  circumscribed  prisms  are  formed 
in  a  triangular  pyramid  whose  base  has  an  area  of  25  sq.  in.  Compute 
the  series  of  differences  between  the  corresponding  sets  when  their  alti- 
tudes are  successively  0.1  in.,  0.01  in.,  0.001  in.,  and  0.0001  in. 

2.  The  base  of  a  pyramid  is  an  equilateral  triangle  3  in.  on  a  side  and 
its  altitude  is  7  in.  Find  the  difference  between  the  corresponding  sets  of 
inscribed  and  circumscribed  prisms  when  the  altitude  of  each  prism  is 
TWTT5  °f  the  altitude  of  the  pyramid. 

3.  Could  a  quadrangular  pyramid  be  used  instead  of  a  triangular 
pyramid  in  the  discussion  of  §§  730-733? 


VOLUMES  OF  PYRAMIDS  AN1>  CONES  359 

736.    Theorem.     The   volumes   of  two    triangular  pyramids, 
having  equivalent  bases  and  equal  altitudes,  are  equivalent. 


Given  the  triangular  pyramids  O-ABC  and  O'-A'B'C ',  having 
equivalent  bases  and  equal  altitudes;  and,  for  convenience, 
having  their  bases  in  the  same  plane. 

To  prove  that  V=V,  where  V  denotes  the  volume  of  O-ABC 
and  V  the  volume  of  O'-A'B'C. 

Proof.  Let  the  common  altitude  ON  be  divided  into  any 
number  of  equal  parts,  and  pass  planes  through  the  points  of 
division  parallel  to  the  plane  of  the  bases.  Construct  the  set 
of  circumscribed  prisms  for  each  pyramid  upon  the  sections 
formed  by  these  planes. 

Then  the  sections  of  the  pyramids  made  by  each  plane  are 
equivalent.  §  710 

Let  DEF  and  D'E'F'  be  the  sections  made  by  any  one  of 
these  planes,  and  P  and  P'  the  corresponding  prisms. 

Then  volume  of  P  =  volume  of  P'.  §  686 

Let  S  denote  the  volume  of  the  sum  of  the  prisms  in  the  set 
of  which  P  is  one;  and  S'  that  of  which  P'  is  one. 

Then  S  =  S'.  §  105 

But  S->7,  and  S'->V.  §  735 

And  S  and  S'  are  variables  that  are  equal  for  all  their  succes- 
sive values  as  the  number  of  the  divisions  of  ON  is  increased 
indefinitely. 

.\V=V.  §485(1) 


S0OCJ: 


'   %SOLID   GEOMETRY 


737.    Theorem.     The  volume  of  a  triangular  pyramid  is  equal 
to  one-third  the  product  of  its  base  and  altitude.     V  =  \Bh. 


Given  the  triangular  pyramid  O-PQR. 

To  prove  that  V  =  \Bh,  where  V  denotes  the  volume  of  the 
pyramid,  B  the  area  of  its  base,  and  h  its  altitude 

Proof.     Upon  the  base  PQR  construct  a  triangular  prism  NR 
of  altitude  h,  and  having  its  lateral  edges  parallel  to  OQ 

The  prism  NR  is  divided  into  three  triangular  pyramids  by 
the  sections  OPR  and  ONR. 

Pyramid  R-NOM  =  pyramid  O-PQR.  §  736 

Pyramid  R-OPQ  =  pyramid  R-ONP.  §  736 

But  pyramid  R-OPQ  is  the  same  as  pyramid  O-PQR. 

Therefore  the  three  triangular  pyramids  are  equal,  and  O-PQR 
is  one-third  the  volume  of  prism  NR. 

But  volume  of  prism  NR  =  Bh.  §  68 

/.  V  =  \Bh.  Why 

738.  Theorem.  The  volume  of  any  pyramid 
is  equal  to  one-third  the  product  of  its  base  and 
altitude.     V  =  \Bh. 

Suggestion.     From  one  vertex  of  the  base 
draw   all    the    diagonals    of    the    base.     Pass  b{ 
planes  through  the  vertex  of  the  pyramid  and 
each  of  these  diagonals.     All  the  pyramids  have    m 
the  same  altitude.     Apply  §  737  to  each  triangular  pyramic 
formed  and  add  the  results. 


VOLUMES  OF  PYRAMIDS  AND  CONES  361 

739.  By  a  like  treatment  to  that  referred  to  in  §  650,  the 
truth  of  the  following  statement  may  be  established: 

The  volume  of  a  circular  cone  is  the  common  limit  of  the  volumes 
of  inscribed  and  circumscribed  pyramids  with  regular  polygons  for 
bases,  as  the  number  of  faces  is  indefinitely  doubled. 

740.  Theorem.  The  volume  of  a  circular  cone  is  equal  to 
one-third  the  product  of  its  base  and  altitude.     V  =  |  Bh  =  \irr2h. 

Given  a  circular  cone. 

To  prove  V =\Bh  =  \irr2h,  where 

V  denotes  volume,  B  area  of  base, 
h  altitude,  and  r  radius. 

Proof.     Inscribe  a  pyramid  with 
a  regular  polygon  for  base,  and  let 

V  denote  its  volume  and  B'  the 
area  of  its  base. 

By  doubling  indefinitely  the  number  of  faces  of  the  pyramid, 
7'->7,  and  B'->B.  §  739,  §  489 

\B'h->\Bh.  §  485  (2) 

But     V  =  \B'h,  being  variables  that  are  always  equal.  §  738 

.\7-jBft.  §485(1) 

Also  B=>irr\  §498 

.-.  F=fnr2fc.  §  111 

Remark.     It  is  to  be  noted  that  the  formulas  of  this  article 
apply  to  any  circular  cone.     Compare  with  the  remark  of  §  719. 

EXERCISES 

37  3V 

1.  Show  that  for  any  pyramid  or  circular  cone  Z?=— ,  and  h  =  ——. 

h  B 

2.  Find  the  volume  of  a  square  pyramid  whose  base  is  16  in.  on  a 
side,  and  whose  altitude  is  10  in. 

3.  Find  the  volume  of  a  pyramid  whose  base  is  a  regular  hexagon 
8  in.  on  a  side,  and  whose  altitude  is  14  in. 

4.  Find  the  volume  of  a  pyramid  whose  base  is  a  square  3\/2  in. 
on  a  side,  and  whose  altitude  is  6f  in. 


362 


SOLID  GEOMETRY 


5.  The  base  edges  of  a  triangular  pyramid  are  6  in.,  8  in.,  and  10  in. 
Find  its  volume  if  the  altitude  is  15  in. 

6.  Find  the  volume  of  a  circular  cone  whose  radius  is  5  in.  and  alti- 
tude 10  in. 

7.  Find  the  volume  of  a  circular  cone  whose  altitude  is  70  ft.  and  the 
circumference  of  whose  base  is  31  ft. 

8.  Derive  a  formula  for  finding  the  radius  of  a  circular  cone  in  terms 
of  the  volume  and  altitude. 

9.  Derive  a  formula  for  finding  the  altitude  of  a  circular  cone  in  terms 
of  the  volume  and  radius. 

10.  A  right  circular  cone  has  a  slant  height  s  and  a  radius  r;  find  its 
volume  V.  Ans.     V  =\lVrWs 

11.  The  Pyramid  of  Cheops  has  a  square  base  720  ft.  on  a  side,  and  an 
altitude  of  480  ft.     Find  the  number  of  cubic  yards  in  it. 

12.  What  is  the  locus  of  the  vertices  of  all  pyramids  having  the  same 
base  and  equal  volumes? 

13.  Find  the  total  area  and  the  volume  of  the  solid  generated  by  revolv- 
ing an  equilateral  traingle  of  side  a  about  one  side.  Find  the  values  if 
a  =  2  in.  Ans.     Area  =  a27T\/3;  21.765  sq.  in, 

Volume  =  Ja37T;   6.283  cu.  in. 

14.  Find  the  total  area  and  the  volume  of  the  solid  generated  by  revolv- 
ing a  parallelogram  with  sides  22  in.  and  16  in.,  and  larger  angle  120°, 
about  one  of  the  longer  sides.       Ans.     3308.3—  sq.  in.;  13270+  cu.  in. 

15.  A  pyramid  whose  base  is  a  square  4  ft.  on  a  side,  and  whose  alti- 
tude is  12  ft.  is  bisected  by  a  plane  parallel  to  the  base.  How  far  from 
the  vertex  is  the  plane? 

16.  A  circular  sheet  of  copper  3  ft.  in  diameter  is  cut  in 
half,  and  each  half  formed  into  a  cone.  These  are  placed 
base  to  base  as  shown  in  the  figure.  This  is  to  be  used  as  a 
float.  Find  the  number  of  pounds  it  will  support  if  the 
sheet  copper  weighs  0.92  lb.  per  square  foot  and  water 
weighs  62.5  per  cubic  foot. 

17.  Hard  coal  dumped  in  a  pile  lies  at  an  angle  of 
30°  with  the  horizontal.  Estimate  the  number  of  tons 
in  a  pile  in  the  shape  of  a  right  circular  cone  having  an 
altitude  of  10  ft.  Large  egg  size  weighs  38  lb.  per 
cubic  foot.  Ans.     About  60  tons. 

18.  Show  how  to  construct  a  triangular  pyramid  equivalent  to  any 
given  pyramid  having  a  polygon  as  base 


VOLUMES  OF  PYRAMIDS  AND  CONES 


363 


741.  Theorem.  The  volume  V  of  a  frustum  of  a  pyramid,  or 
circular  cone,  of  bases  Bx  and  B2  and  altitude  h,  is  given  by  the 
formula  V  =  lh(B1+B2+VB^B'2). 


Given  the  frustum  of  a  pyramid,  or  of  a  circular  cone. 

To  prove  V  =  \h{Bi+B2+  y/BxB2),  where  V  denotes  volume, 
h  altitude,  and  Bi  and  B2  the  areas  of  the  lower  base  and  upper 
base,  respectively. 

Proof.  Complete  the  pyramid,  or  cone,  and  let  m  denote  the 
altitude  of  the  portion  above  the  frustum  in  each  case,  and  V2 
the  volume  of  that  portion.  Also  let  V\  denote  the  volume  of 
the  entire  pyramid,  or  cone. 

Then  V=V1-V2. 

=  iBi(/*+m)-^£2m.  §740 

=  |[51/?,+  (51-52)m]. 

It  remains  to  eliminate  m.  To  do  this  one  more  equation  is 
needed.     This  is  given  by 

Bl_{h+mY 
B2~      m2 

*\fW\     h+m 


709 


Vb2 


m 


Solving  for  m, 
Substituting  this  for  m, 


m  = 


h\/W2 


Vb[-Vb~2 

V^h(B1+B2+VB^2). 


364  SOLID   GEOMETRY 

742.  Theorem.  The  volume  V  of  a  frustum  of  a  circular  cone 
having  bases  with  diameters  d\  and  a\,  and  radii  7*1  and  r%,  and  an 
altitude  hy  is  given  by  the  formulas: 

(1)  V^frrhW+rf+nrt), 

(2)  VmfahW+df+ddti: 

Substitute  irn2  and  7Tr22  for  B\  and  B2  respectively  of  §  741  to  find  (1). 
Use  i7Tdi2  and  lir<k2  for  (2). 

EXERCISES 

1.  Find  the  volume  of  a  frustum  of  a  pyramid  with  square  bases  hav- 
ing edges  of  9  ft.  and  4  ft.  respectively,  and  altitude  12  ft. 

2.  The  radii  of  the  bases  of  a  frustum  of  a  right  circular  cone  are  respec- 
tively 10  cm.  and  12  cm.  Find  its  volume  if  the  altitude  is  16  cm.  Find 
its  lateral  area. 

3.  Find  the  volume  of  a  frustum  of  a  regular  hexagonal  pyramid  the 
bases  being  respectively  8  in.  and  6  in.  on  an  edge,  and  the  altitude  10  in. 

4.  The  frustum  of  a  cone  has  an  altitude  of  8  in.  and  the  radii  of  its 
bases  are  7  in.  and  4  in.  respectively.  Find  its  volume.  Does  the  frus- 
tum have  to  be  cut  from  a  right  cone  in  order  that  its  volume  can  be  found? 

5.  Derive  a  formula  for  finding  the  altitude  of  a  frustum  of  a  circular 
cone  in  terms  of  the  volume  and  the  radii  of  the  bases. 

6.  The  diameter  of  the  top  of  a  water  pail  is  12  in.,  the  diameter  of 
the  bottom  10  in.,  and  the  altitude  10§  in.  How  many  quarts  will  the 
pail  hold?     One  quart  is  57.75  cu.  in.     Ans.     17.33  —  .  K ir 

7.  Find  the  weight  of  an  iron  casting  in  the  form  /— \ 
of  a  right  circular  cone  of  diameter  8  in.  and  altitude           n  ~~\ 

12  in.,  with  a  circular  hole  having  a  diameter  of  2  in.  \      1  \      %, 

through  the  center.     (Cast  iron  weighs  0.26  lb.  per        /    j      '   \ 
cubic  inch.)  Ans.     44.1  lb.      /      (**"*!     V 

8.  A  tank  is  made  in  the  form  of  an  inverted  frus-  L>~ — ! — i"~">\J 

turn  of  a  cone.     The  slant  height  is  14  ft.  and  makes  N • —  ^^S\ 

an  angle  of  60°  with  the  horizontal,  and  the  lower  base  \ 8" H 

is  a  circle  20  ft.  in  diameter.     Find  the  volume  of  the  tank  in  barrels. 
Use  1  bbl.  =4.211  cu.  ft.  Ans.     1685.4  bbl. 

9.  A  tank  is  in  the  form  of  an  inverted  cone  with  its  vertex  angle  60° 
and  its  axis  vertical.  Find  the  depth  of  water  in  the  cone  if  it  has  been 
running  in  for  10  minutes  at  the  rate  of  8  cu.  ft.  per  minute. 


SIMILAR  PYRAMIDS  AND  CONES  365 

SIMILAR   PYRAMIDS   AND   CONES 

743.  Similar  Pyramids.  Pyramids  that  are  formed  from 
the  same  pyramidal  surface  by  parallel  planes  are  similar. 

744.  Similar  Cones.  Cones  that  are  formed  from  the  same 
conical  surface  by  parallel  planes  are  similar. 

745.  Theorem.  The  lateral  areas,  or  the  total  areas,  of  two 
similar  regular  pyramids,  or  right  circular  cones,  are  to  each  other 
as  the  squares  of  two  corresponding  lines,  where  the  corresponding 
lines  may  be  edges,  elements,  altitudes,  slant  heights    or  radii. 

One  case  only  will  be  proved.,  The  other 
cases  are  left  as  exercises. 

Given  two  similar  regular  pyramids 
O-ACDEF  and  O-A'C'D'E'F'. 

S     h2  T      h2 

To  prove — = — ,  and  — = — ,  where  S  and 

S!   h'2        r   h'2 

S'  denote  the  lateral  areas  of  O-ACDEF  and 
O-A'C'B'E'F'  respectively,  h  and  h!  altitudes, 
and  T  and  T'  total  areas. 

Proof.  If  p  and  p'  are  the  perimeters  of  the  respective  bases, 
and  s  and  s'  the  slant  heights,     S  =  ^ps,  and  S'  =  ^p's'.      §  711 

Then  jS=l^i=Z.i.  wh  ? 

S'    %p's'    p'    s' 

S      h2     .         p      h         ,  s      h  w,    9 

.*. —  =  —  since  —  =  — ,  and— =— .  Why: 

S'    h'2  p'    h'  s'    h' 

Further,  if  B  and  B'  are  the  areas  of  the  respective  bases, 

B     h2        _         8     B       ■    8    S'  w,    ? 

—  =  —  and  so  — = — ,  and— = — .  Why: 

B'    h'2  S'    B'9         B    Bf 

Then  ^J?  =  ^±^,  aild  &+*-=*.        §§  406,405 

B  B'  S'+B'    B' 

.-.  — =—  since  S+B  =  T,  and  S'+B'  =  T'.       Why? 
T'    h'2 


366  SOLID  GEOMETRY 

746.  Theorem.  The  volumes  of  two  similar  pyramids,  o 
circular  cones ,  are  to  each  other  as  the  cubes  of  two  correspondim 
lines;  where  the  corresponding  lines  may  J)e  edges,  elements,  alii 
tudes,  slant  heights,  or  radii. 

Using  V  and  V  for  the  respective  volumes,  B  and  B'  for  bases 
and  h  and  h'  for  altitudes, 

Y  _\Bh  _B      h 

V  \B'h'    B'  '  ft7' 

B      h2  V      ft3 

But— =— .     Why?         .-.—  =  —.  Why' 

B'    ft'2  Y'     ft* 

747.  Definition.  The  vertex  angle  of  a  right  circular  com 
is  twice  the  angle  between  an  element  and  the  axis. 

748.  Theorem.  All  right  circular  cones  with  equal  vertex 
angles  are  similar. 

EXERCISES 

1.  Two  cones  generated  by  similar  right  triangles  revolving  abou 
corresponding  sides  are  similar. 

2.  Two  cones  are  generated  by  revolving  two  similar  right  triangle 
about  corresponding  sides  of  length  7  in.  and  10  in.  respectively.  Fine 
the  ratio  of  the  lateral  areas  of  the  cones.  The  ratio  of  the  total  areas 
The  ratio  of  the  volumes. 

3.  The  total  area  of  a  right  circular  cone  is  300  sq.  in.  and  its  altitudi 
is  8  in.  Find  the  total  area  of  the  cone  cut  off  by  a  plane  parallel  to  th4 
base  and  6  in.  from  the  vertex. 

4.  How  far  from  the  vertex  of  a  right  circular  cone,  or  a  regular  pyra 
mid,  of  altitude  h  must  a  plane  be  passed  parallel  to  the  base  so  that  thel 
lateral  area  of  the  part  cut  off  shall  be  one-half  that  of  the  original? 

5.  From  a  given  pyramid,  cut  off  a  frustum  whose  volume  shall  W 
fy  that  of  the  given  pyramid.     Must  the  pyramid  be  regular? 

6.  The  volumes  of  two  pyramids  are  343  cu.  in.  and  1000  cu.  in.  respec- 
tively. The  lateral  area  of  the  smaller  is  100  sq.  in.  Find  the  latera 
area  of  the  larger. 

7.  The  lateral  areas  of  two  similar  right  circular  cones  are  in  the  ratio 
of  9  :  64.  Find  the  volume  of  the  smaller  if  the  volume  of  the  larger  h 
1024  cu.  in. 


QUESTIONS  367 

QUESTIONS 

1.  What  is  a  pyramidal  surface?  A  conical  surface?  Is  the  direc- 
trix necessarily  closed? 

2.  What  are  the  formulas  or  rules  for  finding  the  areas  of  pyramids, 
cones,  and  frustums  of  pyramids  and  cones? 

3.  What  are  the  formulas  or  rules  for  finding  the  volumes  of  each  of 
these  solids? 

4.  Can  you  find  the  area  of  an  oblique  circular  cone?  Of  an  oblique 
pyramid?     Can  you  find  the  volume  of  each  of  these? 

5.  If  the  altitude  of  a  cone  is  divided  into  10  equal  parts  by  planes 
passed  parallel  to  the  base,  how  does  the  area  of  each  section  of  the  cone 

L formed  by  these  planes  compare  with  the  base  of  the  cone? 

6.  Trace  the  steps  in  finding  the  volume  of  a  triangular  pyramid. 
Of  any  pyramid.     Of  a  cone. 

7.  What  is  the  relation  between  two  parallel  sections  of  a  pyramid  or 
cone? 

8.  State  theorems  concerning  the  sections  of  a  cone.     Of  a  pyramid. 

9.  What  effect  does  it  have  upon  the  volume  of  a  pyramid  or  cone  if 
the  area  of  the  base  is  doubled?  If  the  altitude  is  doubled?  If  both 
base  and  altitude  are  doubled? 

10.  What  effect  does  it  have  upon  the  lateral  area  of  a  right  circular 
cone  if  the  circumference  of  the  base  is  doubled?  If  the  area  of  the  base 
is  doubled? 

11.  Point  out  forms  of  this  chapter  that  occur  in  nature.  That  occur 
in  the  arts. 

GENERAL  EXERCISES 

1.  Given  the  following  formulas  for  the  right  circular  cone:  (1)  V  = 
fe-7Tr2/i,  (2)  S  =  Trs,  (3)  S  =  TTrVh*+r*t  (4)  T  =  A+WrVh2+r2,  (5)  T  = 
fTr2+7Tr  V  h2-\-r2)  solve  each  for  each  letter  appearing. 

2.  In  a  right  circular  cone,  A=507T,  and  F  =  2007T.     Findr,  h,  and  s. 

3.  If  R  and  r  are  the  radii  of  the  two  bases  of  a  frustum  of  a  cone, 


Isv 


|;how  that  R+r=\—+Rr. 
\7T/i 

4.    Find  the  volume  and  the  total  area  of  a  right  circular  cone  whose 
ladius  of  base  is  6  in.  and  altitude  is  5.3  in.     Arts.    199.8  cu.  in. ;  263.9  sq.  in. 


368  SOLID  GEOMETRY 

5.  Find  the  weight  of  a  tapered  brick  smoke  stack  175  ft.  high  in  the 
form  of  a  frustum  of  a  right  circular  cone  enclosing  a  cylinder.  The  inner 
diameter  is  10  ft.,  the  wall  is  4  ft.  thick  at  the  base  and  1  ft.  6  in.  at  the 
top.     One  cubic  foot  of  brick  weighs  112  pounds.  Ans.    1095.5  tons. 

6.  Find  the  lateral  edge,  lateral  area,  and  volume  of  a  regular  pyra- 
mid each  side  of  whose  triangular  base  is  5  ft.,  and  whose  altitude  is  9  ft; 

Ans.     9.45  ft.;  68.36  sq.  ft.;  32.5  cu.  ft. 

7.  A  bin  in  a  warehouse  is  12  ft.  square.  A  hopper  is  constructed  on 
the  base,  which  has  a  slope  of  1  :  1.  The  distance  from  the  vertex  of  the 
hopper  to  the  top  of  the  bin  is  18  ft.  Find  the  capacity  of  the  bin  if  t 
bushel  equals  f  cu.  ft. 

8.  The  largest  possible  cylinder  of  diameter  6  in.  is  cut  from  a  right 
circular  cone  having  a  diameter  and  altitude  of  10  in.  and  26  in.  resp< 
tively.     Find  the  volume  of  the  cylinder. 

9.  Find  the  area  and  volume  of  the  solid  generated  hy  revolving 
isosceles  triangle  of  base  10  in.  and  equal  sides  16  in.,  about  its  base.     Fin 
the  area  and  the  volume  of  the  solid  when  this  triangle  is  revolved  about 
one  of  its  equal  sides. 

10.  In  the  frustum  of  a  pyramid  whose  base  is  50  sq.  ft.  and  altitud 
6  ft.,  the  basal  edge  is  to  the  corresponding  top  edge  as  5  to  3.  Find  th 
volume  of  the  frustum.  Ans.     196  cu.  f 

11.  The  base  of  a  pyramid  is  a  rectangle  10  in.  by  8  in.  Find  the  vo 
ume  if  each  lateral  edge  makes  with  the  base  an  angle  of  45°. 

12.  The  total  surface  of  a  regular  quadrangular  pyramid  is  T,  and 
eight  edges  are  equal.     Find  the  length  of  an  edge. 

13.  Find  the  volume  of  a  cone  of  revolution  whose  slant  height  is  equal 
to  the  diameter  of  its  base,  and  whose  total  area  is  T. 

14.  Show  how  to  cut  a  pattern  for  the  frustum  of  a  right  circular  cone, 
if  the  upper  and  lower  bases  have  diameters  of  4  in.  and  6  in.,  respectively, 
and  the  altitude  is  8  in. 

15.  A  right  circular  cone  whose  altitude  is  8  in.  and  radius  6  in.  rolls 
on  a  floor  without  slipping,  making  one  complete  revolution.  What  is  the 
shape  of  the  surface  covered?     Find  its  area. 

16.  A  pyramid  whose  altitude  is  12  in.  weighs  30  pounds.  At  wl 
distance  from  the  vertex  must  it  be  cut  by  a  plane  parallel  to  its  base 
that  the  frustum  cut  off  shall  weigh  20  pounds? 

17.  A  cube  is  cut  by  a  plane  passed  through  the  other  extremities 
the  three  edges  meeting  at  a  vertex.  What  part  of  the  volume  of  the  cul 
is  thus  cut  off? 


CHAPTER  IX 

PRISMATOIDS  AND  POLYEDRONS 

PRISMATOIDS 

749.  A  polyedron  all  of  whose  vertices  lie  in  two  parallel 
planes  is  called  a  prismatoid.  In  either  plane,  the  vertices 
may  lie  in  any  polygon,  as  Fig.  (1);  or,  in  one  plane,  the  ver- 
tices may  lie  in  a  line,  as  in  Fig.  (2) ;  or  there  may  be  a  point 
only,  as  in  a  pyramid.     The  faces  whose  vertices  are  in  the 


parallel   planes   are   evidently  triangles,    or   quadrilaterals   of 
types  that  have  at  least  two  parallel  sides.     Why? 

750.    The  altitude  of  a  prismatoid  is  the  distance  between 
the  two  parallel  planes. 

EXERCISES 

1.  Show  that  a  prism  is  a  prismatoid  whose  bases  are  congruent  poly- 
gons. 

2.  Show  that  a  frustum  of  a  pyramid  is  a  prismatoid  whose  bases  are 
similar  polygons. 

3.  Draw  or  describe  a  prismatoid  that  has  a  face  that  is  a  trapezoid. 
A  parallelogram.     A  rectangle. 

4.  Compute  the  area  of  the  midsection  of  a  pyramid  whose  base  is  a 
regular  hexagon  having  a  side  of  6  in. 

6.    Compute  the  areas  of  the  midsection  of  the  prismatoid  in  Fig.  (2)  if 
ABCD  is  a  rectangle  having  A£  =  18  in.  and  2?C  =  11  in.,  and  EF  =  12  in. 

369 


370 


SOLID  GEOMETRY 


N, 


751.  Theorem.  If  Bi  and  B2  are  the  areas  of  the  two  bases, 
M  the  area  of  the  midsection,  and  h  the  altitude,  then  the  volume 
V  of  a  prismatoid  is  given  by  the  formula  V  =  ^h(Bi+B2+4:M). 

Given  a  prismatoid  of  volume  V,  alti- 
tude h,  bases  Bi  and  B2,  and  midsection  M. 

To  prove  V  =  ih(B1+B2+4M). 

Proof.     If  any  lateral  face  is  not  a  tri- 
angle divide  the  face  into  triangles  by  a  E 
diagonal. 

Take  any  point  0  in  the  midsection  and^- 
connect  it  with  the  vertices  of  the  prisma- 
toid and  the  midsection.  These  lines 
form  triangles  with  the  edges  of  the  prismatoid,  which  separate 
the  prismatoid  into  pyramids  that  have  the  bases  and  faces  of 
the  prismatoid  as  bases. 

(1)  Pyramid  0-ABCD=lhBv  §737 

(2)  Pyramid  0-NPQ  =  \hB2. 
Each  of  the  pyramids  like  O-CDP,  which  may  be 

called  a  lateral  pyramid,  can  have  its  volume  ex- 
pressed in  terms  of  the  part  of  the  midsection  com- 
mon to  it.  To  prove  this  consider  O-CDP  apart 
from  the  prismatoid,  and  draw  HD. 

Altitude  of  each  of  pyramids  P-OHI  and  D-OHI  is  \h 

Pyramid  P-OHI  =  \h  times  area  OHI.  §   737 

Pyramid  D-OHI  -  \h  times  area  OHI.  Why? 

Pyramid  O-CDH  =  2  times  pyramid  0-HDL  Why? 

Then  pyramid  O-CDH  =  %h  times  area  OHI. 

Hence  pyramid  O-CDP  =  %h  times  area  OHI.  Why? 

Similarly  each  lateral  pyramid  is  equal  to  -f  h  times  the  area 
of  its  part  of  the  midsection. 

(3)  Therefore   the   sum   of   the   lateral   pyramids  =  j^hM  = 


\h-±M. 

Adding  (1),  (2), 


Why? 


and  (3), 

7  =  |fc(B1+B2+4M). 


PRISM  ATOIDS  371 

752.    Other  Applications  of  the  Prismatoid  Formula.     The 

formula  for  the  volume  of  a  prismatoid  can  be  applied  also  to 
various  other  solids  such  as  cylinders,  cones,  spheres,  combi- 
nations of  these,  and  various  other  forms  that  cannot  well  be 
described  here. 

The  prismatoid  formula,  because  of  its  wide  application,  is 
used  often  by  engineers  and  others  in  determining  the  volumes 
of  embankments,  abutments,  and  other  irregular  solids  of  vari- 
ous kinds. 

EXERCISES 

1.  Derive  the  following  formulas  from  the  formula  for  the  volume  of  a 
prismatoid : 

(1)  Prism  and  cylinder,  V  =  Bh. 

(2)  Pyramid  and  cone,  V  =  ^Bh. 

(3)  Frustum  of  pyramid  or  cone,  V = %h  {Bi  +B2  +  \ZBiB2)  . 
Derivation  for  frustum. 

Given  V=*§h(Bi+Bs+4M). 

To  derive  V  =  \h{Bl  +B2  +  y/BiB2). 

It  is  then  necessary  to  express  M  in  terms  of  B\  and  B2. 

Using  the  notation  of  §  741,  — =-7T- — ,  and  —  =  711 T~-        §  709 

M     (J/i+ra)2  M     $h+m)2 

_    V#i      h+m         .  \/B2         m 
Or     , — =y~ ,  and- 


/ M     \h-\-m          \/M     f/i+ra' 
Solving  each  of  these  for  m,  ra  = j= — 2  ; —  —  from  the  first: 

Vb[-Vm 

and  m  —  — j== y=  from  the  second. 


Vm-Vb, 

h(VM-§y/Bi)_     hhVW2 

Vb[-Vm      Vm-VK 

Dividing  by  to^Sr^ 


Then 


VBi-VM     Vm-Vb2 

a/M  a/M 

Then2VM-V^^Vg'  ^  ^M  =  VW1  +  VB,  §  406 

Or  4M  =  £1+£2+2V£1£2. 
.'.  V  =ih(2Bl+2B2+2\/BA)  =^(Bi+B,  +  Vb,B,). 


372 


SOLID  GEOMETRY 


2.  Show  that,  in  general,  the  midsection  of  a  prismatoid  has  as  many 
sides  as  the  two  bases  combined.     When  will  this  not  be  so? 

3.  Find  the  volume  of  the  frustum  of  a  square  pyramid,  the  lower  base 
being  8  ft.  square,  the  upper  base  6  ft.  square,  and  the  altitude  12  ft.  Solve 
both  by  the  frustum  formula  and  the  prismatoid  formula,  and  compare. 

4.  Both  bases  of  a  prismatoid  of  altitude  h  are  squares,  and  the  lateral 
faces  isosceles  triangles.  The  sides  of  the  upper  base  are  parallel  to  the 
diagonals  of  the  lower  base  and  half  the  length  of  these  diagonals.  If  b  is 
a  side  of  the  lower  base,  find  the  volume.     Ans.    %b2h. 

5.  Find  the  volume  of  the  prismatoid  with  dimen- 
sions as  shown  in  the  figure.  The  bases  are  right  tri- 
angles   having    their    corresponding    sides    parallel. 

Ans.     3420  cu.  in. 

6.  Use  the  prismatoid  rule  to  find  the  volume  of  a 
frustum  of  a  pyramid  whose  bases  are  regular  hexagons 
10  in.  and  6  in.  on  a  side  respectively,  and  whose  altitude  is  18  in. 

7.  Find  the  volume  of  the  frustum  of  the  preceding  exercise  by  §  741 
and  compare  the  result  with  that  of  the  preceding.  n  ^ 

8.  The  volume  of  a  truncated  triangular  prism  is 
readily  determined  by  the  prismatoid  formula.  To  do 
this  consider  one  face  and  the  opposite  edges  as  the 
bases. 

Find  the  volume  of  the  truncated  triangular  prism 
shown  in  the  figure.  The  base  ABC,  which  has  a  right 
angle  at  C,  is  a  right  section. 

9.  Show  that  the  volume  of  any  truncated  triangular  prism  is  equal  to 
the  product  of  the  area  of  a  right  section  and  D  |" 
one-third  the  sum  of  the  three  edges. 

Suggestion.  Let  the  edges  be  c,  d,  and  e, 
and  the  altitude  and  base  of  a  right  section  a 
and  b  respectively. 

Consider  the  prism  as  a  prismatoid  with 
bases  ABED  and  CF,  and  midsection  M.       A 

Then  7  =  £a[i&(c+d)+0+4M].  §751 

But  M  =  j6-i[i(c+e)+|(d+c)]. 

•*•  V —  \ab'  J(c+d-j-e)=area  of  right  section  X \ (c-j-d+e). 

10.  A  truncated  right  triangular  prism  has  for  base  an  isosceles  triangle 
whose  sides  are  10  in.,  10  in.,  and  8  in.  Find  its  volume  if  the  three 
lateral  edges  are  5  in.,  7  in.,  and  11  in.  respectively. 


POLYEDRONS 


373 


POLYEDRONS 

753.  Regular  Polyedrons.  A  polyedron  whose  faces  are 
congruent  regular  polygons  v  and  whose  polyedral  angles  are 
congruent  is  called  a  regular  polyedron. 

A  polyedron  of  four  faces  is  a  tetraedron;  one  of  six  faces, 
a  hexaedron;  one  of  eight  faces,  an  octaedron;  one  of  twelve 
faces,  a  dodecaedron;  and  one  of  twenty  faces,  an  icosaedron. 


It  is  instructive  to  construct  the  regular  polyedrons  as  shown 
in  the  following  figures.  Draw  on  cardboard  the  diagrams 
shown  below,  cut  along  the  full  lines,  and  fold  along  the  dotted 
lines.  The  edges  that  meet  may  be  fastened  with  gummed 
paper.  Durable  models  may  be  constructed  from  tin  and 
soldered  at  the  edges. 

There  are  many  interesting  facts  connected  with  the  five  regular  poly- 
edrons. The  so-called  fourteenth,  fifteenth,  and  sixteenth  books  of  Euclid 
give  many  theorems  and  problems  concerning  these  solids. 


374 


SOLID   GEOMETRY 


754.  Theorem.  There  can  be  no  more  than  five  regular  poly- 
edrons. 

The  proof  of  this  theorem  depends  upon  the  facts: 

(a)  At  least  three  planes  must  meet  to  form  a  polyedral  angle.    (§  603). 

(b)  The  sum  of  the  face  angles  of  a  polyedral  angle  is  less  than  360°. 
(§  609). 

(c)  The  faces  of  a  regular  polyedron  are  congruent  regular  polygons. 

(1)  Faces  equilateral  triangles. 

A  polyedral  angle  can  be  formed  of  three,  four,  or  five  equilateral 
triangles,  and  no  more.  Why? 

Therefore  no  more  than  three  regular  polyedrons  can  be  formed  having 
equilateral  triangles  as  faces. 

(2)  Faces  squares. 

A  polyedral  angle  can  be  formed  of  three  squares,  and  no  more.        Why? 
Therefore  no  more  than  one  regular  polyedron  can  be  formed  having 
squares  for  faces. 

(3)  Faces  regular  pentagons. 

A  polyedral  angle  can  be  formed  of  three  regular  pentagons,  and  no 
more.  Why? 

Therefore  no  more  than  one  regular  polyedron  can  be  formed  having 
regular  pentagons  for  faces. 

Regular  polygons  with  a  greater  number  of  sides  than  five  cannot  be 
used  to  form  polyedral  angles  because  the  sum  of  three  or  more  angles  of 
such  polygons  is  equal  to  or  greater  than  360°. 

Hence  there  can  be  no  more  than  five  regular  polyedrons. 

755.  Relation  between  the  Number  of  Faces,  Edges,  and 
Vertices  of  a  Regular  Polyedron.  It  is  interesting  to  note  the 
relation  between  the  number  of  faces,  edges,  and  vertices  of  a 
regular  polyedron.  Complete  the  following  table,  and  com- 
pare with  the  table  on  page  67. 


Name 

Number 

of 

Faces 

Form  of 
Faces 

Number 

of 
Edges 

Number 

of 
Vertices 

Number 
of  Faces 
at  Each 
Vertex 

Sum  of 

Face 

Angles 

Tetraedron 

4 

Equi.  A 

6 

4 

3 

720° 

Hexaedron 

Octaedron 

Dodecaedron 

Icosaedron 

POLYEDRONS  375 

756.  Euler's  Theorem.  This  theorem  is  stated  because  of 
its  interest.  The  proof  is  too  difficult  to  be  given  here.  As 
an  exercise  it  may  be  tested  for  the  regular  polyedrons.  Use 
the  preceding  table. 

Theorem.  In  any  polyedron  the  number  of  edges  plus  two  is 
equal  to  the  sum  of  the  number  of  faces  and  the  number  of 
vertices. 

Note.  Leonhard  Euler  was  born  in  Switzerland  in  1707  and  died  in 
Petrograd  in  1783.  He  was  one  of  the  greatest  physicists,  astronomers, 
and  mathematicians  of  the  eighteenth  century. 

757.  Theorem.  The  sum  of  the  face  angles  of  any  polyedron 
is  equal  to  360°  multiplied  by  two  less  than  the  number  of  ver- 
tices. 

The  proof  of  this  theorem  is  based  upon  the  preceding  theorem.  Test 
it  for  the  regular  polyedrons. 

EXERCISES 

1.  Find  the  area  of  the  surface  of  a  regular  tetraedron  3  in.  on  an  edge. 
Of  a  regular  octaedron  4  in.  on  an  edge.  Of  a  regular  icosaedron  6  in.  on 
an  edge.  Ans.     15.588+  sq.  in.;  55.426—  sq.  in.;  311.769+  sq.  in. 

2.  Find  the  area  of  the  surface  of  a  regular  dodecaedron  2  in.  on  an 
edge.  Ans.  82.583— sq.  in. 

3.  If  the  edge  of  a  regular  tetraedron  is  e,  show  that  its  volume  is  given 
by  the  formula  V  =  Y^e3\/r2. 

Solution.     V = %ON  X  area  of  AABC.  J* 

DC  =  %e\/3.  See  p.  166,  Ex.  7  (3).  //I\ 

Area  of  AABC  =  \AB  X  ZX7  =  Je2  y/l,  /  \\     \ 

OC  =  JDC  =  \eVz.        ^_  _      §211.    PZ      /|___AC 

ON  =  VcN2  -OC2  =  V>  -  (p  V3)2  -  e  Vf .  dQ3^ 

...  7=JeV/fxie2V3=A-^V/2.  J 

4.  If  the  edge  of  a  regular  octaedron  is  e,  show  that  its  volume  is  given 
by  the  formula  V =\ez\f2. 

5.  Compare  the  volumes  of  a  regular  tetraedron,  a  cube,  and  a  regular 
octaedron,  each  2  in.  on  an  edge. 

6.  Find  the  area  and  volume  of  a  regular  tetraedron  having  an  altitude 
of  8  in. 


376  SOLID  GEOMETRY 

758.  Theorem.  The  volume  of  two  tetraedrons  that  have  a 
triedral  angle  of  one  congruent  to  a  triedral  angle  of  the  other  are 
to  each  other  as  the  products  of  the  three  edges  of  these  triedral 
angles.  ct 


L >JS' 


Given  the  tetraedrons  O-ABC  and  O'-A'B'C,  with  triedral 
ZO  =  triedral  ZO' . 

„  V        OA-OB-OC         u        T7       .  ry,   , 

To  prove— =___-____,  where  V  and  V  denote  the 

volumes  of  the  tetraedrons. 

Proof.  Place  tetraedron  O-ABC  so  that  triedral  ZO  will 
coincide  with  triedral  ZO'. 

Draw  CN  and  CM  perpendicular  to  O'A'B'. 

CN  and  CM  determine  a  plane  which  intersects  O'A'B'  in 

O'M.  Why? 

AT  V      O'AB-CN       O'AB      CN 

Now  — = —  = .  Why? 

V    O'A'B' -CM    O'A'B'    CM 

But  VA^^O'A-0'B^ 

O'A'B'    O'A'-O'B' 
Further        AO'NC  and  AO'MC  are  similar  rt.A.  Why? 

Then  M-^..  1428 

CM    O'C 


V      O'A-0'B-O'C         OA-OB-OC 


§111 


V    O'A'-O'B' -O'C    O'A'-O'B' -O'C 
Compare  this  theorem  and  the  proof  with  §  375. 

759.  Theorem.  Two  parallelepipeds  thai  have  a  triedral 
angle  of  one  congruent  to  a  triedral  angle  of  the  other  are  to  each 
other  as  the  products  of  the  three  edges  of  these  triedral  angles. 


POLYEDRONS 


377 


760.  Similar  Polyedrons.  Two  polyedrons  are  similar  if 
they  have  the  same  number  of  faces  similar  each  to  each  and 
similarly  placed,  and  have 
their  corresponding  poly- 
edral  angles  congruent.  It 
is  evident  that  the  corre- 
sponding diedral  angles 
are  also  equal. 

761.  Theroem.  The  cor- 
responding edges  of  similar 
polyedrons  are  proportional. 

Corresponding  edges  belonging  to  corresponding  faces  are  proportional 
by  §  444.  Use  the  fact  that  each  edge  belongs  to  two  faces  and  prove  the 
theorem. 

762.  Theorem.  The  surfaces  of  two  similar  polyedrons  are  to 
each  other  as  the  squares  of  any  two  corresponding  edges. 

Given  two  similar  polyedrons  P  and  P\  with  corresponding 
faces  A  and  A',  B  and  Bf,  C  and  C,  •  •  • .  Also  given  a  and  a', 
b  and  &',  c  and  c',  •  •  • ,  corresponding  sides  of  these  faces  respec- 
tively. 


_               A+B+C+--- 
To  prove 

A'+B'  +  C'+- 
corresponding  edges. 

A      a2    B 


=  — ,  where  n  and  n'  are  any  two 


n ' 


Proof. 


But 


And  hence 


Therefore 


=  ¥    C 
A'    an-' B'    bf '  C' 
a     b     c 

a'~b'~7'~ 
a2  _  ¥  _  &  _ 

A'    B'    C'~ 
■    A+B+C+--- 
"  A'+B'+C'+-- 


§446 

Why? 

Why? 

§403 


378  SOLID  GEOMETRY 

763.  Theorem.  The  volumes  of  two  similar  tetraedrons  are  to 
each  other  as  the  cubes  of  any  two  corresponding  edges. 

Use   §§  746,  761. 

764.  Theorem.  The  volumes  of  two  similar  polyedrons  are  to 
each  other  as  the  cubes  of  any  two  corresponding  edges. 

This  theorem  will  not  be  proved.  It  is  accepted  as  true  since  it  has 
useful  applications. 

EXERCISES 

1.  What  is  the  ratio  of  the  volumes  of  two  cubes  that  are  5  in.  and 
4  in.  on  an  edge?     What  is  the  ratio  of  their  areas? 

2.  The  edges  of  a  parallelepiped  are  10,  12,  and  14.  Find  the  edges 
of  a  similar  parallelepiped  having  ^  as  great  an  area.  Find  the  ratio  of 
their  volumes. 

3.  Two  pyramids  are  cut  from  the  same  pyramidal  surface.  The 
lateral  edges  of  one  are  9  in.,  12  in.,  and  14  in.  and  of  the  other  18  in.,  21 
in.,  and  20  in.     Find  the  ratio  of  their  volumes. 

4.  Find  the  ratio  of  the  volume  of  a  regular  octaedron  6  in.  on  an 
edge  to  the  volume  of  a  regular  octaedron  having  half  as  great  an  area. 

5.  The  center  of  gravity  of  a  tetraedron  is  J  its  altitude  above  the 
base.     Find  the  center  of  gravity  of  a  regular  tetraedron  8  in.  on  an  edge. 

6.  The  edge  of  a  regular  tetraedron  is  e.  Find  the  edge  of  a  regular 
tetraedron  that  has  a  volume  n  times  as  great  as  the  volume  of  the  given 
tetraedron. 

7.  Find  the  edge  of  a  regular  tetraedron  such  that  its  volume  multi- 
plied by  \/2  is  288. 

8.  Find  the  volume  of  a  regular  tetraedron  that  has  an  altitude  of  17  in. 

9.  A  section  of  a  tetraedron  by  a  plane  parallel  to  two  opposite  edges 
is  a  parallelogram. 

10.  The  midpoints  of  the  edges  of  a  regular  tetraedron  are  the  vertices 
of  a  regular  octaedron. 

11.  Two  similar  polyedrons  have  volumes  of  121.5  cu.  in.  and  4.5  cu. 
in.,  respectively.  An  edge  of  the  smaller  is  1 J  in.  Find  the  corresponding 
edge  of  the  larger.  Ans.     4.5  in. 

12.  The  area  of  the  entire  surface  of  a  polyedron  is  108  sq.  in.,  and  its 
volume  is  432  cu.  in.  If  the  area  of  the  entire  surface  of  a  similar  polyedron 
is  75  sq.  in.,  find  its  volume.  Ans.     250  cu.  in. 


GENERAL  EXERCISES 


379 


GENERAL  EXERCISES 


U—iot—i 


1.  If  from  any  point  within  a  regular  tetraedron  perpendiculars  are 
drawn  to  its  faces,  their  sum  equals  an  altitude  of  the  tetraedron. 
(Compare  this  with  Ex.  2,  p.  96.) 

2.  A  wedge  whose  altitude  is  10  in.  and  edge  4 
in.,  has  a  base  that  is  a  square  having  a  perimeter 
of  36  in.     Find  the  volume  of  the  wedge. 

3.  A  concrete  pier  for  a  railway  bridge  has  dimen- 
sions as  shown  in  the  figure,  the  bases  being  rec- 
tangles with  semicircles.  Find  the  number  of  cubic 
yards  of  concrete  in  the  pier.     Ans.     37.8—  cu.  yd. 

4.  A  railroad  cut  has  the 
dimensions  given  in  the  figure, 
which  shows  the  vertical  sec- 
tion and  three  cross  sections, 
one  at  each  end  and  one  in  the 
middle.  Find  the  number  of 
cubic  yards  of  earth  removed  in 
digging  the  cut. 

Ans.     7972f  cu.  yd. 
Suggestion.     The  part  on  each  side  of  the  center  is  a  prismatoid,  and 
can  be  computed  separately. 

5.  A  tank  of  reinforced  concrete  is  160  ft.  long,  100  ft.  wide,  and  10  ft. 
6  in.  deep,  outside  dimensions.  The  side  walls  are  8  in.  thick  at  the  top 
and  18  in.  thick  at  the  bottom,  with  the  slope  on  the  inside.  The  bottom 
is  6  in.  thick.  Find  the  number  of  cubic  yards  of  cement  used  in  building 
the  tank,  and  the  capacity  of  the  tank  in  gallons. 

6.  The  vertices  of  one  regular  tetraedron  are  the  centers  of  the  faces  of 
another  regular  tetraedron.  Find  the  ratio  of  the  volumes  of  the  two 
tetraedrons. 

7.  The  three  planes  passing  through  the  lateral  edges  of  a  triangular 
pyramid,  and  bisecting  the  base  edges,  meet  in  a  common  straight  line. 

8.  The  six  planes  passing  through  the  edges  of  a  tetraedron,  and 
bisecting  the  opposite  edges,  meet  in  a  common  point. 

9.  The  point  of  intersection  of  the  planes  in  the  preceding  exercise  is 
the  center  of  gravity  of  the  tetraedron.  Prove  that  the  center  of  gravity 
of  a  tetraedron  divides  the  line  from  a  vertex  to  the  center  of  gravity 
of  the  opposite  face  in  parts  that  are  in  the  ratio  of  3  :  1. 

The  center  of  gravity  of  a  /ace  is  at  the  intersection  of  the  medians 
of  that  face. 


380 


SOLID  GEOMETRY 


10.  The  corresponding  edges  of  three  similar  tctraedrons  are  3  in 
4  in.,  and  5  in.,  respectively.  Find  the  corresponding  edge  of  a  simila 
tetraedron  equal  in  volume  to  their  sum. 

11.  By  means  of  the  prismatoid  formula  show  that  the  volume  of 
truncated  quadrangular  prism  whose  opposite  faces 
are  parallel,  is  equal  to  the  area  of  a  right  section 
times  one-fourth  the  sum  of  the  lateral  edges. 

Suggestion.  Consider  two  of  the  parallel  faces, 
as  Bi  and  B2,  as  bases.  Then  M  is  the  midsection 
between  these. 

Let  A  be  the  area  of  a  right  section  having  a 
side  in  Bi  equal  to  n,  and  altitude  h,  the  distance 
between  Bi  and  B2. 

Then £i=4  (a-f-d)  n, -'&«■'$.  (6+c)  n,  and M ~§ { \  (a+b)+%(c+d)  }n 
'•  V-\h  \\  (a+d)  n+J  (6+c)  *+2{|  (a+&)+|  {c+d)}n] 
=  \hn  (a+&+c+d)=jA  (a+b+c+d). 

12.  The  bottom  of  a  bin  of  wheat  is  a  rectangle  5  ft.  by  12  ft.  The 
top  of  the  wheat  is  in  a  plane  such  that  the  depths  at  the  corners  are 
6  ft.,  5  ft.,  3  ft.,  and  4  ft.  respectively.  Find  the  number  of  bushels  in 
the  bin  if  1  bu.  =  l|  cu.  ft. 


x"1 — 

i 

i 

7 

I 

M  1 

i 
i 

Bi 
b 

,'--- 
'" 

CHAPTER  X 
THE  SPHERE  AND   POLYEDRAL  ANGLES 


GENERAL   PROPERTIES 

765.  Sphere.  A  sphere  is  a  closed  surface  all  points  of 
which  are  equally  distant  from  a  point  within,  called  the 
center  of  the  sphere. 

A  sphere  is  usually  designated  by  a  single  letter  at  its  center. 

The  radius  of' a  sphere  is  the  straight  line  connecting  the 
center  to  a  point  in  the  surface.  A  straight  line  connecting 
two  points  in  the  surface  and  also  passing  through  the  center 
is  a  diameter. 


A  sphere  separates  space  into  two  parts  so  that  any  point 
that  does  not  lie  in  the  surface  lies  within  the  sphere  or  outside  it. 

The  distance  from  the  center  to  a  point  within  a  sphere  is 
less  than  the  length  of  the  radius;  while  a  point  outside  a  sphere 
is  at  a  greater  distance  from  the  center  than  the  length  of  the 
radius. 

766.  Great  Circle  of  a  Sphere.  It  is  evident  from  the 
definitions  of  a  sphere  and  of  a  circle  that  a  plane  passing 
through  the  center  of  sphere  intersects  the  sphere  in  a  circle 
ivhose  radius  is  equal  to  the  radius  of  the  sphere.  Such  a 
rircle  i-s  called  a  great  circle  of  the  sphere. 

381 


382  SOLID  GEOMETRY 

767.  The  following  facts  follow  readily  from  the  definitions 
given: 

(1)  Radii  of  the  same  sphere,  or  equal  spheres,  are  equal. 

(2)  Diameters  of  the  same  sphere,  or  equal  spheres,  are  equal 

(3)  Spheres  having  equal  radii,  or  equal  diameters,  are  equal 

(4)  All  great  circles  of  a  sphere  are  equal. 

(5)  The  points  of  intersection  of  any  two  great  circles  of  a  sphen 
are  in  a  diameter. 

(6)  A  great  circle  of  a  sphere  bisects  the  sphere. 

(7)  A  sphere  may  be  generated  by  the  revolution  of  a  semicircU 
about  Us  diameter. 

768.  The  Straight  Line  and  the  Sphere.    In  order  to  stud^ 
the  relative  positions  of  a  straight  line  and  a  sphere,  pass  i 
plane  through  the  line  and  the  center  of  the         S*     "\ 
sphere.     The  line  will  then  have  the  same     /p="^-~~3m2f 
relations  to  the  sphere  that  it  bears  to  the  ^jl^t-       -^j\ 
great  circle  in  this  plane.  ^^_^^ 

(1)  If  the  line  intersects  the  great  circle  it  cuts  the  circU 
in  two  points,  and  likewise  the  sphere. 

(2)  If  the  line  is  tangent  to  the  great  circle  it  has  only  om 
point  in  common  with  the  sphere,  and  is  perpendicular  to  the 
radius  at  this  point. 

769.  Definition.  If  a  line  has  but  one  point  in  commoi 
with  the  sphere  it  is  said  to  be  tangent  to  the  sphere. 

EXERCISES 

1.  A  line  perpendicular  to  a  radius  at  its  extremity  is  tangent  to  th 
sphere. 

2.  The  locus  of  all  lines  tangent  to  a  sphere  at  a  point  is  a  plane  per 
pendicular  to  a  radius  at  its  outer  extremity. 

3.  Find  the  locus  of  the  centers  of  all  spheres  tangent  to  a  given  lin 
at  a  given  point. 

4.  Find  the  locus  oi  the  centers  of  all  spheres  of  a  given  radius  tangen 
to  a  fixed  line. 


GENERAL  PROPERTIES  OF  SPHERE  383 

5.  The  plane  perpendicular  to  a  tangent  at  its  point  of  contact  passes 
through  the  center  of  the  sphere. 

6.  What  is  the  locus  of  the  vertices  of  the  right  angles  of  the  right 
triangles  having  a  given  hypotenuse? 

770.  The  Plane  and  the  Sphere.  If  a  plane  has  but  one 
point  in  common  with  a  sphere,  it  is  said  to  be  tangent  to 
the  sphere. 

771.  Theorem.  A  plane  perpendicular  to  a  radius  of  a  sphere 
at  its  outer  extremity  is  tangent  to  the  sphere.  ^^_^^ 

Given  plane  P  perpendicular  to  the          /                 >v 
radius  OC  at  its  outer  extremity  C.  r~~- -A 

To  prove  plane  P  is  tangent  to  the        fc"       ~a       ~~^y\ 

sphere  0.  r\       /  ■  J~\ 

Give  a  proof  similar  to  that  of  §  284.     /      \j<a  •  c  -J/      \ 

fp    b^ ^  \ 

772.  Theorem.     A  plane  tangent  to  a 

sphere  is  perpendicular  to  the  radius  drawn  to  the  point  of  contact. 

For  if  the  plane  was  not  perpendicular  to  the  radius  it  would  have  a 
point  less  than  the  length  of  the  radius  from  the  center  of  the  sphere. 

773.  Theorem.  If  a  plane  is  tangent  to  a  sphere,  the  radius 
drawn  to  the  point  of  tangency  is  perpendicular  to  the  plane. 

EXERCISES 

1.  Two  lines  perpendicular  to  a  radius  at  its  outer  extremity  deter- 
mine a  plane  tangent  to  a  sphere. 

2.  Planes  tangent  to  a  sphere  at  the  extremities"  of  a  diameter  are 
parallel. 

3.  Explain  how  to  construct  a  line  tangent  to  a  sphere  at  a  point  on 
ttlbhe  sphere.     Explain  how  to  construct  a  tangent  plane  at  this  point. 

4.  Explain  how  to  construct  a  line  and  a  plane  tangent  to  a  sphere 
3ff|ind  through  an  external  point. 

5.  A  point  B  is  22  in.  from  the  center  0  of  a  sphere  having  a  radius 
liffibf  12  in.     Find  the  distance  from  B  to  C,  the  point  of  tangency  of  a  plane 

;hrough  B. 

6.  Find  the  locus  of  the  centers  of  all  spheres  tangent  to  two,  giveja 
lit ersec ting  planes. 


384  SOLID  GEOMETRY 

CIRCLES   OF   SPHERES 

774.    Theorem.     The  section  of  a  sphere  made  by  a  plane  is  c 
circle,  whose  center  is  the  foot  of  the  perpendicular  from  th 
center  of  the  sphere  to  the  plane. 

Given  the  sphere  0  cut  by  the  plane  P  in 
the  section  ABN,  also  OQ  perpendicular  to 
plane  P. 

To  prove  that  section  ABN  is  a  circle 
with  center  Q. 

Proof.  Draw  QA  and  QB  to  any  two  points  A  and  B  on  t 
section  ABN.     Also  draw  OA  and  OB. 

AAQO  and  ABQO  are  congruent  right  triangles.  Whyl 

Then  AQ  =  BQ.     That  is,  any  two  points  on  section  ABN  aflj 

equidistant  from  Q.  Whyl 

.*.  section  ABN  is  a  circle  with  center  Q.  Whyl 

EXERCISES 

1.  If  a  plane  that  intersects  a  sphere  is  gradually  moved  from  til 
center,  describe  the  change  in  the  intersection  of  the  plane  with  the  sphere 

2.  A  line  that  is  tangent  to  a  sphere  is  tangent  to  each  section  of  j 
sphere  formed  by  a  plane  containing  the  line. 

3.  The  radius  of  a  sphere  is  14  in.  Find  the  area  of  a  section  ma 
by  a  plane  10  in.  from  the  center. 

4.  The  radius  of  a  sphere  is  12  in.  If  the  area  of  a  section  made  by 
plane  is  314.16  sq.  in.,  find  the  distance  from  the  center  of  the  sphere  1 
the  plane. 

5.  How  far  from  the  center  of  a  sphere,  having  a  radius  of  12  in.,  mul 
a  plane  pass  so  that  the  section  made  by  this  plane  shall  be  one-half  thi 
area  of  a  great  circle  of  the  sphere? 

6.  In  the  same  sphere  or  equal  spheres,  two  sections  equally  distan 
from  the  center,  are  equal,  and  conversely.     Compare  with  §  281. 

7.  What  is  the  locus  of  the  projections  of  a  given  point  upon  the  pla 
passing  through  another  point? 

8.  What  is  the  locus  of  a  point  from  which  three  planes  can  be  dra 
tangent  to  a  given  sphere  and  form  a  triedral  angle  all  of  whose  die 
angles  are  right? 


CIRCLES  OF  SPHERES  385 

775.  Definitions.     A  small  circle  of  a  sphere  is  the  inter- 
section  of   the   sphere   with   any  plane  not 
passing  through  its  center. 

The  axis  of  a  circle  of  a  sphere  is  the 
diameter  of  the  sphere  that  is  perpendicular 
to  the  plane  of  the  circle. 

The  poles  of  a  circle  of  a  sphere  are  the 
extremities  of  the  axis  of  the  circle. 

In  the  figure,  AMB  is  a  great  circle,  CND  is  a  small  circle,  PP'  is  the 
axis  of  each  of  these  circles,  and  P  and  P'  are  the  poles. 

776.  Theorem.  Through  any  three  points  on  a  sphere  one  and 
only  one  circle  of  the  sphere  can  be  drawn. 

For  the  three  points  determine  a  plane. 

777.  Theorem.  Through  any  two  points  on  a  sphere  a  great 
circle  can  be  drawn. 

What  third  point  is  in  the  plane  of  the  great  circle? 
How  are  the  points  situated  when  only  one  great  circle  can  be  drawn? 
How  when  more  than  one  can  be  drawn? 

778.  Theorem.  Parallel  circles  of  a  sphere  have  the  same  axis 
and  the  same  poles. 

779.  Theorem.     //  one  great  circle  of  a  sphere 
is  perpendicular  to  another ,  either  circle  contains  A 
the  axis  and  poles  of  the  other;  and  conversely. 

Use  §§  767  (5),  594. 

EXERCISES 

1.  If  the  surface  of  the  earth  is  considered  a  sphere,  what  kind  of 
circles  are  the  equator,  the  meridians,  and  the  parallels  of  latitude?  What 
is  the  axis  of  the  equator?  Of  the  parallels  of  latitude?  What  are  the 
poles  of  each  of  these? 

2.  Find  the  circumference  of  the  parallel  of  latitude  45°  north  of  the 
equator.  Of  the  parallel  of  latitude  60°  north.  (Use  4000  miles  as  the 
radius  of  the  earth.) 


386  SOLID  GEOMETRY 

DISTANCE   ON   A   SPHERE 

780.  By  the  distance  between  two  points  on  a  sphere  is 
meant  the  length  of  the  minor  arc  (§  264)  of  the  great  circle 
through  the  two  points.  That  this  is  the  shortest  path  on 
the  sphere  between  the  two  points  can  be  proved. 

If  the  two  points  are  at  the  extremities  of  a  diameter,  the 
distance  between  them  is  a  semicircumference  of  the,  great 
circle. 

It  should  be  noted  that  an  arc  of  a  great  circle  of  a  sphen 
here  takes  the  place  of  a  straight  line  in  a  plane  in  determining 
the  distance.     That  is,  the  distance  between  two  points  in  a 
plane  is  measured  along  the  straight  line  joining  them,  whiL 
on  a  sphere  the  distance  is  measured  along  a  great  circle. 

781.  A  quadrant  of  a  sphere  is  one  quarter  of  the  circum- 
ference of  a  great  circle  of  the  sphere. 

782.  Polar  Distance.  The  polar  distance  of  a  small  circle 
of  a  sphere  is  the  distance  on  the  sphere  from  the  nearer  pole 
to  a  point  in  the  circle. 

The  polar  distance  of  a  great  circle  may  be  taken  from  eithe: 
of  its  poles  to  a  point  in  the  circle. 

783.  Theorem.  The  polar  distances  of  a  circle  of  a  sphere 
are  equal. 

Given  OO'  on  sphere  0  with  poles  N  and  S.     /~'7<\ 

To  prove  that  all  points  in  OO'  are  equally  /  A/ 
distant  from  N,  or  S. 

Proof.     Take  A  and  B  any  two  points  in  Y  \ 
OO',  and  draw  great  circles  through  NAS  and      N\ 
NBS;  also  draw  lines  AO\  BO',  AN,  and  BN. 

Prove  AAO'N^ABO'N,  and  therefore  AN  =  BN. 

Then  AN  =  BN. 

In  all  points  in  OO'  are  equally  distant  from  N, 

,\  like  manner  prove  for  pole  & 


DISTANCE  ON  A  SPHERE  387 

784.  Theorem.  On  the  same  sphere  or  on  equal  spheres,  the 
polar  distances  of  equal  circles  are  equal. 

785.  Theorem.  The  polar  distance  of  a  great  circle  is  a  quad- 
rant; and  conversely. 

786.  Theorem.  The  straight  lines  joining  the  points  in  the 
circle  of  a  sphere  to  a  pole  are  equal. 

EXERCISES 

1.  What  is  the  name  of  the  circle  on  the  surface  of  the  earth  at  a 
quadrant's  distance  from  the  north  pole?  At  23^°  from  the  north  pole? 
At  23^°  from  the  south  pole?     At  23^°  from  the  equator? 

2.  Show  how  to  construct  a  circle  on  a  sphere  and  at  a  given  distance 
from  a  pole.     Discuss  for  a  small  circle  and  for  a  great  circle. 

3.  Show  how  to  construct  a  circle  of  given  radius  on  the  surface  of 
a  sphere. 

4.  If  the  radius  of  a  sphere  is  10  in.,  find  the  radius  of  a  circle  having  a 
polar  distance  of  45°.     Of  30°. 

787.  Theorem.  If  a  point  on  a  sphere  is  at  a  quadrant's  dis- 
tance from  each  of  two  other  points,  not  the  extremities  of  a  diam- 
eter, on  the  sphere,  it  is  a  pole  of  the  great  circle  through  these 
two  points.  ,  n 

Given  a  point  TV  on  a  sphere  0,  at  a  quad-   ff        ^^\ 

rant's  distance  from  each  of  points  A  and  B,  I    L — | \\ 

and  ABC  the  great  circle  through  A  and  B.    rH""^~"J}Jc 

To  prove  that  N  is  the  pole  of  QABC.         W      j y 

Proof.  ZAON  and  ZBON  are  rt.A.  Why?   \^Jl/ 
Then    ON  is  perpendicular   to    plane    of  s 

QABC.  §  571 

And  SN  is  the  axis  of  QABC.  Why? 

.'.  N  is  a  pole  of  QABC.  Why? 

EXERCISES 

1.  Show  how  to  find  the  pole  of  a  great  circle  on  a  sphere. 

2.  Show  how  to  construct  a  great  circle  through  two  points  on  a 
sphere. 


388  SOLID  GEOMETRY 

788.  Circumscribed  Polyedrons.  A  polyedron  is  said  to  be 
circumscribed  about  a  sphere  when  the  sphere  is  tangent  to 
each  face  of  the  polyedron.  The 
sphere  is  then  said  to  be  inscribed 
in  the  polyedron. 


/■'■'    !fl;;V\ 


789.  Inscribed    Polyedrons.      A 

polyedron  is  said  to  be  inscribed  in  a  /  m  «  \ 

sphere  when  all  its  vertices  lie  in  the        / 

sphere.     The  sphere  is  then  said  to  be 

circumscribed  about  the  polyedron.       ^L-~ " 

790.  Theorem.  One,  and  only  one,  sphere  can  be  inscribed  in 
a  given  tetraedron.  jj> 

Given  the  tetraedron  ABCD.  / \  \. 

To  prove  that  one  sphere  and  only  one       /EA  \. 

can  be  inscribed  in  ABCD.  a<~-— U --^>c 

Outline  of  proof.     Construct  the  planes       ^\^^^^ 
bisecting  the  diedral  angles  AD  and  BD.  B 

These  planes  will  intersect  in  a  line  DS.  Why? 

Then  DS  is  the  locus  of  all  points  equally  distant  from  faces 
ADC,  ABD,  and  BDC.  Why? 

Construct  the  plane  bisecting  the  diedral  angle  AB. 

This  plane  will  intersect  DS  in  some  point  0.  Why? 

Then  0  is  equally  distant  from  the  four  faces  of  the  tetrae- 
dron. Why? 

If  a  sphere  is  constructed  with  0  as  center  and  the  perpen- 
dicular from  0  to  any  face,  as  OE,  for  radius,  this  sphere  will 
be  tangent  to  each  face  of  the  tetraedron,  and  therefore  is  an 
inscribed  sphere.  §  789 

To  show  that  only  one  inscribed  sphere  can  be  constructed 
show  that  there  is  only  one  point  0. 

Compare  this  theorem  and  proof  with  the  corresponding 
problem  in  plane  geometry.     (§  325). 

Exercise.  Find  the  locus  of  the  centers  of  all  spheres  tangent  to  the 
faces  of  a  triedral  angle. 


SPHERES  AND  TETRAEDRONS  389 

791.  Theorem.  One,  and  only  one,  sphere  can  be  circum- 
scribed about  a  given  tetraedron. 

Given  the  tetraedron  ABCD. 

To  prove  that  one  sphere  and  only  one  can 
be  circumscribed  about  ABCD. 

Outline  of  proof.  Circumscribe  a  circle 
about  AABC,  and  erect  a  perpendicular  O'Q 
to  its  plane  at  its  center  0'. 

Then  O'Q  is  the  locus  of  all  points  equally  distant  from  A, 
B,  and  C.  Why? 

At  the  midpoint  E  of  the  edge  CD  construct  a  plane  P  per- 
pendicular to  CD. 

Then  plane  P  is  the  locus  of  all  points  equally  distant  from 
C  and  D.  Why? 

But  plane  P  intersects  O'Q  in  a  point  0.  Why? 

Then  0  is  equally  distant  from  the  four  vertices  of  the  tet- 
raedron, and  a  sphere  having  0  as  center  and  OA  as  radius  will 
pass  through  the  four  vertices  A,  B,  C,  and  D.  Why? 

Therefore  a  sphere  can  be  circumscribed  about  the  tetraedron. 

To  show  that  only  one  sphere  can  be  circumscribed  about 
ABCD,  show  that  there  is  only  point  0. 

Compare  this  theorem  and  proof  with  the  corresponding 
problem  in  plane  geometry.     (■§§  316,  317). 

EXERCISES 

1.  What  is  the  locus  of  points  equidistant  from  four  points  not  all  in 
the  same  plane? 

2.  How  many  points  are  necessary  to  determine  a  sphere?  How 
many  points  on  a  sphere  are  necessary  to  determine  it  if  its  center  is  given? 

3.  Prove  that  the  planes  that  are  the  perpendicular  bisectors  of  the 
edges  of  a  tetraedron  pass  through  a  common  point. 

4.  Prove  that,  if  a  circle  is  circumscribed  about  each  face  of  a  tetra- 
edron, the  perpendiculars  to  the  respective  faces  at  the  centers  of  the 
circumscribing  circles  are  concurrent. 

5.  One,  and  only  one,  sphere  can  be  inscribed  in  or  circumscribed 
about  a  cube. 


390 


SOLID   GEOMETRY 


792.    Problem.     To  find  the  radius  of  a  given  material  sphere. 


Given  a  material  sphere  0. 

To  find  its  radius. 

Construction.  Choose  any  point  N  on  the  sphere  as  a  pole, 
and  describe  a  convenient  circle  ABC. 

Choose  any  three  points  on  this  circle  as  A,  B,  and  C. 

Construct  in  a  plane  AA'B'C  congruent  to  AABC.  §  25 

Circumscribe  a  circle  with  center  Q'  about  AA'B'C. 

Draw  Q"B"  equal  to  the  radius  Q'B',  and  then  construct 
XY±  Q"B"  and  through  Q". 

With  the  compasses  lay  off  B"N'  equal  to  BN,  to  meet  XY 
at  N',  and  construct  B"S'  _L  B"N'  at  B"  and  extend  it  to  meet 
XY  at  S'. 

Determine  the  bisecting  point  0'  of  S'N'. 

Then  O'N'  is  the  radius  of  the  sphere  0. 

The  proof  is  left  to  the  student. 


EXERCISES 

1.  If  NB  =  13  in.  and  QB  =  12  in.,  compute  ON. 

2.  By  means  of  an  instrument  called  a  sphero- 
meter,  the  distances  QN  and  QB  can  be  measured. 
Show  how  the  radius  ON  can  then  be  computed. 
Such  an  instrument  is  used  practically  in  deter- 
mining the  radius  of  curvature  of  a  spherical  lens. 

3.  Find  the  radius  of  curvature  of  a  spherical 
lens,  that  is,  the  radius  of  the  sphere  of  which  the 
surface  of  the  lens  is  a  part,  if  QN  =0.125  cm.  and 
QB  m  2  cm.    If  QN  -  0.643  cm,  and  QB  m  2.5  cm, 


Arts.     16.9  in. 


SPHERES,  CYLINDERS,   CONES  391 

793.  Sphere  Inscribed  in  Cylinder  or  Cone.  A  sphere  is 
said  to  be  inscribed  in  a  cylinder,  or  the  cylinder  circum- 
scribed about  the  sphere,  when  the  bases  of  the  cylinder  and 
all  its  elements  are  tangent  to  the  sphere. 

A  sphere  is  said  to  be  inscribed  in  a  cone,  or  the  cone  cir- 
cumscribed about  the  sphere,  when  the  base  of  the  cone  and 
all  its  elements  are  tangent  to  the  sphere. 

794.  Cylinder  or  Cone  Inscribed  in  a  Sphere.  A  cylinder 
is  said  to  be  inscribed  in  a  sphere,  or  the  sphere  circum- 
scribed about  the  cylinder,  when  its  bases  are  circles  of  the 
sphere. 

A  cone  is  said  to  be  inscribed  in  a  sphere,  or  the  sphere 
circumscribed  about  the  cone,  when  its  base  is  a  circle  of  the 
sphere  and  its  vertex  lies  on  the  sphere. 

795.  Theorem.  All  the  tangents  to  a  sphere  from  a  given 
point  are  equal,  and  their  points  of  contact  are  in  a  circle  of  the 
sphere. 

Given  sphere  0,  and  point  P  outside  the 
sphere. 

To  prove  that  the  tangents  to  the  sphere 
from  the  point  P  are  equal,  and  that  their 
points  of  contact  lie  in  a  circle. 

Outline  of  proof.     Let  PB  be  one  tangent. 
Then  PB  is  tangent  to  a  great  circle  determined  by  P,  B, 
and  0.  Why? 

Therefore  ZPBO  is  a  rt.Z.  §  288 

Pass  a  plane  through  B  _L  PO.  This  will  intersect  the  sphere 
in  a  circle  with  center  0' '. 

Further,  aline  drawn  from  P  to  any  point,  as  C,  in  the  QOf 
is  tangent  to  the  sphere  0,  for  AOCP  =  AOBP,  and  therefore 
PC  J_  OC.  Why? 

Show  that  no  tangent  from  P  could  have  its  point  of  contact 
outside  of  OO'. 

That  the  tangents  are  all  equal  follows  from  §  584. 


392 


SOLID  GEOMETRY 


have 


a  common 


RELATIVE  POSITIONS   OF   SPHERES 

796.  The  relative  positions  of  two  spheres  are  analogous  to 
the  relative  positions  of  two  circles.     (See  §§  291-296.) 

797.  Line  of  Centers.  The  line  joining  the  centers  of  two 
spheres  is  the  line  of  centers. 

798.  Tangent  Spheres.     If   two    spheres    are    tangent    to 
the  same  plane  at  the  same  point,  they  are 
tangent   spheres.     They  may  be  tangent 
internally  or  externally. 

Spheres  A  and  B  are  tangent  internally;  and  A 
and  C  externally. 

799.  Concentric   Spheres.     Spheres  that 
center  are  concentric  spheres. 

800.  Theorem.  The  intersection  of  two  spheres  is  a  circle, 
whose  center  is  in  a  straight  line  joining  the  centers  of  the  spheres 
and  whose  plane  is  perpendicular  to  that  line. 

Given  two  intersecting  spheres  0  and 
0'. 

To  prove  that  the  intersection  of  the 
spheres  is  a  circle  whose  center  is  in  00' 
and  whose  plane  is  perpendicular  to  00' . 

Outline  of  proof.  Through  0,  0' ,  and  any  point  A  of  the 
intersection,  pass  a  plane.  This  plane  intersects  the  two 
spheres  in  two  great  circles  intersecting  in  two  points.       Why? 

If  A  and  B  are  these  points,  then  00'  is  the  perpendicular 
bisector  of  AB.  §  295 

Show  that  if  the  entire  figure  is  revolved  about  00',  the  great 
circles  generate  the  spheres,  and  point  A  generates  a  circle  with 
center  C  and  radius  CA,  which  is  common  to  the  spheres. 

Also  show  that  the  plane  of  the  circle  generated  by  A  is  per- 
pendicular to  00'. 

Further,  show  that  if  any  point  of  the  intersection  was  out- 
side of  this  circle,  then  the  spheres  would  coincide.     (§  791). 


MEASUREMENT  OF  THE  SPHERE,  AREA  393 

801.  Theorem.  //  two  spheres  are  tangent  to  each  other ,  the 
line  of  centers  passes  through  the  point  of  contact. 

Proof  similar  to  that  of  §  296. 

EXERCISES 

1.  State  and  prove  exercises  concerning  the  sphere  analogous  to  Exer- 
cises 1  to  5,  page  115. 

2.  What  is  the  locus  of  points  at  a  distance  r  from  a  given  point  O, 
and  at  a  distance  r'  from  a  given  point  0"? 

3.  Find  the  center  of  a  sphere  which  contains  a  given  circle  and  also 
contains  a  given  point  not  lying  in  the  plane  of  the  circle. 

4.  Two  spheres  with  radii  34  and  50,  respectively,  intersect.  If  the 
distance  between  their  centers  is  56,  find  the  radius  of  their  circle  of 
intersection,  and  the  distance  from  the  center  of  each  sphere  to  the  center 
of  the  circle  of  intersection. 

5.  The  line  of  contact  of  a  sphere  inscribed  in  a  circular  cone  with 
the  conical  surface  of  the  cone,  is  a  small  circle  of  the  sphere. 

6.  The  line  of  contact  of  a  sphere  inscribed  in  a  circular  cylinder  with 
the  cylindrical  surface,  is  a  great  circle  of  the  sphere. 

MEASUREMENT   OF   THE   SPHERE,   AREA 

802.  Area  of  a  Sphere.  By  the  area  of  a  sphere  is  meant 
the  numerical  measure  of  the  surface,  that  is,  the  number  of 
units  of  area  in  it. 

As  with  the  cylinder  and  cone,  a  sphere  is  a  curved  surface, 
and  a  plane  unit  of  area  cannot  be  applied  directly  to  it  to  find 
its  numerical  measure.  Its  area  can  be  determined  as  a  limit. 
It  is  based  upon  (7)  of  §  767,  which  states  that  a 
sphere  is  generated  by  revolving  a  semicircle  about 
its  diameter,  and  the  following  statement  which  is 
accepted  without  proof: 

If  in  a  semicircle  generating  a  sphere,  one-half  of  a 
regular  polygon  be  inscribed  so  that  a  vertex  lies  at 
each  end  of  the  diameter,  then,  as  the  number  of  the  sides  of  the 
inscribed  semipolygon  is  indefinitely  doubled,  the  area  generated 
by  the  semipolygon  approaches  the  area  of  the  sphere  as  a  limit. 


394 


SOLID  GEOMETRY 


803.  Theorem.  The  area  of  the  surface  generated  by  a  seg- 
ment of  a  straight  line  revolving  about  an  axis,  in  its  plane  but  not 
perpendicular  to  it  and  not  intersecting  it,  is  equal  to  the  projection 
of  the  segment  upon  the  axis  multiplied  by  the  circumference  of 
the  circle  whose  radius  is  the  perpendicular  erected  at  the  midvoint 
of  the  segment  and  terminated  by  the  axis. 


Given  the  segment  AB  revolving  about  the  axis  XY  in  its 
plane. 

To  prove  that  S  =  00'X2wXEF,  where  S  denotes  the  area  of 
the  surface  generated,  00'  is  the  projection  of  AB  upon  XY, 
and  FE  is  the  perpendicular  at  the  midpoint  of  AB. 

Proof.     Three  cases  arise: 

Case  I,  in  which  AB  is  oblique  to  and  does  not  meet  XY. 
The  surface  generated  is  the  lateral  surface  of  a  frustum  of  a 
right  circular  cone. 

The  proof  of  this  is  given  in  §  729. 

Case  II,  in  which  A B  is  oblique  to  and  meets  XY.  The  sur- 
face generated  is  the  lateral  surface  of  a  right  circular  cone. 

S  =  irXOAxAB.  §719 

Show  that  ADEF^AOAB. 

Then    AB  :  OB  =  EF  :  DF,  and  ABXDF  =  OBX  EF. 

Since   OB  =  00'  and  OA  =  2DF,  ABxOA=  200' X EF. 

.'.S  =  00'X2ttXEF.  §  111 

Case  III,  in  which  AB  is  parallel  to  XY.  The  area  generated 
is  the  lateral  surface  of  a  right  circular  cylinder. 

Give  proof. 


MEASUREMENT  OF  THE  SPHERE,  AREA  395 

804.    Theorem.     The  area  of  a  sphere  is  equal  to  four  times 
the  area  of  a  great  circle  of  the  sphere.    S  =  47rr2. 


Given  the  area  S  of  a  sphere  generated  by  the  revolution  of 
the  semicircle  0  of  radius  r,  about  the  diameter  AE. 

To  prove  that  £=47rr2. 

Proof.  Inscribe  in  the  semicircle  one-half  of  a  regular  poly- 
gon of  an  even  number  of  sides,  as  ABCDE.  Let  A  denote  the 
apothem  of  the  polygon,  and  let  F,  0,  and  G  be  the  projections 
of  B,  C,  and  D  respectively  upon  the  diameter  AE. 

For  each  side,  the  apothem  is  the  perpendicular  bisector.  Why? 

Then  area  generated  by  AB  =  AF X  2tcl.  §  803 

Similarly      area  generated  by  BC  =  FOX2ira. 

Etc. 

Adding  these  equations  and  denoting  by  S'  the  area  generated 
by  the  revolution  of  the  semipolygon  ABCDE, 
Sf=(AF+FO+'-)2ira. 

But  (AF+FO+---)=AE  =  2r. 

Then  £'  =  47rra,  which  is  always  true  as  the  number  of  the 
sides  of  the  polygon  is  continuously  doubled. 

Further  S'->S  and  since  a->r,  ^wra^wr2.  §§  802, 490,  485  (2) 

/.  S=47rr2.  §485(1) 

Show  that  other  forms  of  the  formula  for  the  area  of  a  sphere 
are  S  =  2irrdJ  and  S=7rd2. 

805.    Theorem.     The  areas  of  two  spheres  are  to  each  other  as 
the  squares  of  their  radii;  or,  as  the  squares  of  their  diameters. 
For,  if  S,  and  S'  are  the  areas  of  two  spheres, 

_^  =  j^l!  =  l!_  and—  =  —  =  — 
S'    4:wr'2    r,2>  ^    S'~ ird'2~ d'2 


396  SOLID  GEOMETRY 

EXERCISES 

1.  Given  the  area  of  a  sphere,  to  find  its  radius  and  its  diameter. 

2.  Find  the  area  of  a  sphere  whose  radius  is  5.  Of  a  sphere  whose 
diameter  is  16.     Of  a  sphere  whose  circumference  is  24  in. 

3.  Find  the  radius  of  a  sphere  whose  area  is  100  sq.  in. 

4.  How  many  square  feet  of  tin  will  it  take  to  roof  a  hemispherical 
dome  40  ft.  in  diameter? 

5.  Find  the  diameter  of  a  sphere  that  has  the  same  area  as  a  cube 
that  is  8  in.  on  an  edge. 

6.  Find  the  area  of  a  sphere  that  is  circumscribed  about  a  cube  that 
is  12  V3  in.  on  an  edge.  Ans.     4071.5  sq.  in. 

7.  The  number  of  square  inches  in  the  area  of  a  sphere  is  equal  to 
the  number  of  linear  inches  in  the  circumference  of  a  great  circle  of  the 
sphere.     Find  the  radius  of  the  sphere. 

8.  Show  that,  if  a  cylinder  is  circumscribed  about  a 
sphere,  the  lateral  area  of  the  cylinder  is  equal  to  the  area 
of  the  sphere.  Also  show  that  the  area  of  the  sphere  equals 
two-thirds  the  total  area  of  cylinder. 

9.  The  radius  of  the  earth  is  3960  miles  and  the  radius 
of  the  moon  is  1080  miles.     Find  the  ratio  of  their  areas. 

10.  Show  how  to  determine  the  distance  apart  the  points  of  a  compass 
must  be  placed  to  draw  a  great  circle  upon  a  material  sphere  that  has 
a  diameter  of  14  in.     So  as  to  draw  a  circle  having  a  radius  of  5  in. 

11.  What  is  the  locus  of  the  centers  of  all  spheres  tangent  to  a  given 
plane  at  a  given  point? 

12.  What  is  the  locus  of  the  centers  of  all  spheres  passing  through 
three  given  points? 

13.  What  is  the  locus  of  points  at  a  distance  n  from  a  point  Oi  and  at 
a  distance  r2  from  another  point  O2?     Discuss. 

14.  Find  the  area  of  a  sphere  circumscribed  about  a  tetraedron  whose 
edge  is  4  in. 

15.  The  radii  of  two  intersecting  spheres  are  10  in.  and  12  in.,  respec- 
tively, and  the  distance  between  their  centers  is  16  in.  Find  the  radius 
of  the  circle  of  intersection  of  the  two  spheres. 

16.  Three  equal  spheres  each  tangent  to  the  other  two  rest  on  a  plane. 
A  fourth  sphere  rests  on  the  first  three  spheres.  Find  the  distance  from 
the  center  of  the  fourth  sphere  to  the  plane,  if  each  sphere  has  a  radius 
of  8  in.  Ans.     21.064  in. 


MEASUREMENT  OF  THE  SPHERE,  AREA  397 

806.  Zones.  The  portion  of  the  surface  of  a  sphere  in- 
cluded between  two  parallel  planes  is  called         . 

a  zone.  /^SS^\\ 

It  is  evident  that  a  zone  can  be  consid-  4/x^_  1Z^>\ 
ered  as  generated  by  an  arc  of  a  great  y  5  T1 
circle.  A,  >A 

The    circles    made    by    the    two    parallel   L — ^ — A 

planes  are  called  the  bases  of  the  zone,,  and  the  distance 
between  the  planes  is  the  altitude  of  the  zone. 

If  one  of  the  planes  is  tangent  to  the  sphere  the  zone  is 
called  a  zone  of  one  base. 

The  zones  on  the  surface  of  the  earth  are  included  between  certain 
parallels  of  latitude.  The  torrid  and  temperate  zones  are  zones  of  two 
bases,  and  the  frigid  zones  are  zones  of  one  base. 

807.  Theorem.  The  area  of  a  zone  of  a  sphere  is  equal  to  the 
product  of  the  altitude  of  the  zone  and  the  circumference  of  a  great 
circle  of  the  sphere;  that  is,  area  =  2wrh,  where  r  denotes  the  radius 
of  a  great  circle  and  h  the  altitude  of  the  zone. 

The  proof  is  similar  to  that  of  §  804. 

808.  Theorem.  The  areas  of  zones  of  the  same  sphere  or  of 
equal  spheres  are  to  each  other  as  their  altitudes. 

EXERCISES 

1.  Find  the  area  of  a  zone  of  one  base  and  having  an  altitude  of  5  ft. 
on  a  sphere  8  ft.  in  diameter. 

2.  Find  the  radius  of  a  sphere,  on  which  a  zone  of  altitude  5  in.  has  an 
area  of  100  sq.  in. 

3.  The  diameter  of  a  sphere  is  12  in.,  and  is  divided  into  four  equal 
parts  by  parallel  planes.     Find  the  area  of  each  zone. 

4.  Find  what  part  of  the  earth's  surface  is  north  of  the  parallel  45° 
north.     North  of  the  parallel  60°  north. 

5.  Show  that  one-half  of  the  earth's  surface  lies  within  30°  of  the 
equator. 

6.  What  part  of  the  earth's  surface  could  be  seen  from  a  point  whose 
distance  from  the  surface  is  equal  to  the  radius?     Twice  the  radius? 


398  SOLID  GEOMETRY 

809.  Lunes.    That  portion  of  a  sphere  between  the  halves 
of  two  great  circles  of  the  sphere  is  called  a        ^jf^ 
lune. 

The  diedral  angle  between  the  planes  of  the  c{^_|_|q_|!^2> 
great  circles  bounding  the  lune  is  the  angle  of 
the  lune. 

From  §  589  it  follows  that  the  angle  of  a  lune  is  measured 
by  the  plane  angle  of  the  diedral  angle  formed  by  the  planes 
of  the  great  circles  bounding  the  lune. 

810.  Comparison  of  Lunes.  If,  on  the  same  sphere,  one 
lune  is  placed  on  another  with  one  bounding  semicircle  in 
common,  the  other  bounding  semicircles  either  coincide,  or 
one  lies  wholly  within  the  lune  bounded  by  the  other.  It  is 
evident  then  that: 

(1)  On  the  same  sphere  or  equal  spheres ,  two  lunes  having  equal 
angles  are  equal;  and  conversely. 

(2)  On  the  same  sphere  or  equal  spheres,  two  lunes  are  in  the 
same  ratio  as  their  angles. 

811.  Theorem.     The  area  of  a  lune  is  given  by  the  formula 

u 

L  = (47rr2),  where  L  denotes  the  area  of  the  lune,  u  the  anqle  of 

360  . 

the  lune  in  degrees,  and  r  the  radius  of  the  sphere. 

To  prove  this,  consider  the  sphere  as  a  lune  whose  angle  is 

360°,  then  by  §  810  (2),  — =  — ,  or  L  =  — (47rr2). 
4?rr2     360  360 

EXERCISES 

1.  Find  the  area  of  a  lune  whose  angle  is  30°  on  a  sphere  8  ft.  in 
diameter. 

2.  Find  the  part  of  the  earth's  surface  between  the  equator  and  the 
parallel  30°  north,  and  between  the  meridians  120°  west  and  150°  west. 

3.  The  chord  of  the  polar  distance  of  a  circle  is  8  in.  If  the  radius 
of  the  sphere  is  10  in.,  what  is  the  radius  of  the  circle?  What  is  the  area 
of  the  zone  cut  off  by  the  circle? 


MEASUREMENT  OF  THE  SPHERE,  AREA 


399 


4.  Find  the  altitude  of  a  zone  whose  area  equals  the  area  of  a  great 
circle  of  the  sphere. 

5.  A  plane  passes  through  a  sphere  bisecting  at  right  angles  a  radius 
of  the  sphere.     Compare  the  areas  of  the  two  parts  of  the  sphere. 

6.  If  h  is  the  height  of  P  above  the  surface  of  the  earth  and  r  the  radius, 

then  S,  the  extent  of  the  surface  of  the  earth  visible  from  P,  is  given  by 

2lTr2h 
the  formula    S  =  - 


r+h 
S  =  2TrXMQ. 
But  MQ=r-OM,  and  OM  :  OA=OA  :  OP. 


§807 


Then  OM  = -,  and MQ=r- 

r+h 


~r+h 


rh 

'r+h' 


:.S  =  2TTrX- 


rh       2lTr2h 


r-\-h      r-\-h 

7.  How  many  miles  above  the  surface  of  the  earth  is  a  point  from 
which  one-fourth  of  the  surface  can  be  seen?  Arts.     r. 

8.  ,  From  a  point  6  ft.  from  a  sphere  one-fourth  of  its  area  is  visible. 
Find  the  radius  of  the  sphere. 

9.  An  ash  tray  is  in  the  form  of  the  zone  of  a  sphere.  If  the  tray  is 
1  in.  deep  and  3^  in.  in  diameter  at  the  top,  find  the  diameter  of  the  blank 
from  which  it  is  pressed.  C 

10.  A  zone  of  one  base  has  the  same  area  as 
a  circle  whose  radius  is  the  chord  of  the  gener-  Ai 
ating  arc  of  the  zone,  that  is,  the  area  of  the  D 

zone  generated  by  arc  BC  revolving  about  DC  is  equal  in  area  to  a  circle 
having  chord  BC  for  a  radius. 

11.  Show  that  the  area  of  a  zone  of  one  base  is  given  by  the  formula 

S  =  lT(w2  +  W)  =7T[(|w)2+/i2], 
where  S  denotes  the  area,  w  the  diameter  of  the  base  of  the  zone  and  h 
the  altitude. 

12.  In  a  practical  handbook,  the  following  rule  is 
given  as  nearly  correct.  Show  that  it  is  accurate. 
The  area  of  a  flanged  spherical  segment  is  equal  to  a 
circle  of  radius  equal  in  length  to  the  fine  drawn  from 
the  top  of  the  segment  to  the  edge  of  the  flange,  that  is 
equal  to  a  circle  of  radius  AB. 

Suggestion.    Area  of  zone  =  7r[(|^)2+/i2]. 

Area  of  flange  =  7T[(^)2-(^)2]._ 
Area  of  zone  and  flange  =  7T [ (^d)2+h2]=TT A B2. 


400 


SOLID   GEOMETRY 


VOLUME   OF  A   SPHERE 

812.  By  the  volume  of  a  sphere  is  meant  the  volume 
within  the  sphere. 

813.  If  a  regular  polyedron,  for  instance  a  cube,  is  circum- 
scribed about  a  sphere,  and  lines  drawn  from  each  vertex  to  the 
center  of  the  sphere,  these  lines  may  be  taken  as  the  lateral 
edges  of  pyramids  having  their  vertices  at  the  center  of  the 
sphere  and  having  the  faces  of  the  polyedron  as  bases.  The 
volume  of  the  polyedron  can  then  be  expressed  as  the  volume 
of  these  pyramids. 


Further,  if  tangent  planes  be  drawn  where  each  lateral  edge 
intersects  the  sphere,  a  circumscribed  polyedron  of  a  greater 
number  of  faces  will  be  formed.  It  is  evident  that  this  process 
can  be  continued  indefinitely,  thus  forming  polyedrons  whose 
areas  and  volumes  steadily  decrease,  and  can  be  made  to  differ 
as  little  as  desired  from  the  area  and  volume,  respectively,  of 
the  sphere.     The  following  statement  can  then  be  accepted. 

814.  If  each  vertex  of  a  circumscribed  polyedron  is  joined  to  the 
center  of  the  sphere,  and  tangent  planes  are  drawn  at  the  points 
where  these  lines  intersect  the  sphere,  then,  as  the  number  of  faces 
is  increased  indefinitely,  the  area  and  the  volume  of  the  polyedron 
approach  as  limits  the  area  and  the  volume,  respectively,  of  the 
sphere. 


VOLUME   OF   A  SPHERE 


401 


815.    Theorem.     The  volume  of  a  sphere  is  equal  to  the 
product  of  its  area  by  one-third  of  its  radius. 

H 


N 


jot 

Given  a  sphere  0  of  radius  r  and  area  S. 

To  prove  that  the  volume  V  =  SX^r. 

Proof.  Circumscribe  about  the  sphere  any  polyedron,  for 
instance  a  cube,  and  denote  the  surface  of  the  polyedron  by  S'. 

Since  the  polyedron  may  be  considered  as  composed  of  pyra- 
mids having  the  faces  of  the  polyedron  as  bases  and  the  center 
of  the  sphere  as  common  vertex,  the  volume  V  of  the  poly- 


edron is 


V'  =  S'xb 


If  the  number  of  faces  of  the  polyedron  is  indefinitely  in- 
creased as  described  in  §  813, 

f->V,  S'->S,  and  S'x\r->Sx\r.     §§  814,  485  (2) 

But  Vf  =  SfX^r  is  always  true  as  the  number  of  faces  is 
increased. 


v=sxh 


r. 


§485 

816.    Theorem.     The  volume  V  of  a  sphere  of  radius  r  and 
diameter  d  is  given  by  the  formulas:    V  =  j^Trrz  and  V  =  \irdz. 
For  V  =  \rS  by  §  815.     But  >S  =  4?rr2  or  ird?  by  §  804. 


V  =  4irr 


3  and  V  =  \ird\ 


817.  Theorem.  The  volumes  of  two  spheres  are  to  each  other 
as  the  cubes  of  their  radii,  or  as  the  cubes  of  their  diameters. 

Let  Vi,  rh  di  and  Fi,  r2,  a\  be  the  volumes,  radii  and,  diam- 
eters, respectively. 


Then 


Ii=i^l=rZ  0YYi=b^l^ 


fjrr2* 


r2* 


V2    Indf    a\* 


402 


SOLID   GEOMETRY 


818.  Theorem.     The  prismoid  formula  holds  for  a  sphere. 

In  the  formula,  V  =  %h(Bi+B2+4M)  of  §  751,  /i  =  2r,  Bi  =  Q, 
52  =  0,  4M  =  47rr2.     ,\  7  =  |7rr3. 

819.  Theorem.     The  volume  of  a  sphere  is  equal  to  two-thirds 
the  volume  of  the  circumscribed  right  cylinder. 


EXERCISES 

1.  Consider  the  surface  of  a  sphere  divided  as  shown 
in  the  figure.  Can  you  conclude  that  the  volume  is  as 
given  in  §  815? 

2.  Show  that  the  volumes  of  the  right  circular  cylinder, 
the  sphere,  and  the  right  circular  cone  with  dimensions  as 
given  in   the  figure,    are   in 


the  ratio   of   3:2:1. 

Note.  Archimedes  (287- 
212  b.c.)  who  was  the  great- 
est mathematician  of  antiq- 
uity, proved  the  theorem  of 
Exercise  2.  This  theorem 
may  well  be  regarded  as  one  of  the  most  beautiful  in  elementary  mathe- 
matics. Archimedes  proved  many  of  the  most  important  theorems  in 
regard  to  the  sphere. 

3.  State  a  formula  for  finding  the  radius  of  a  sphere  when  the  volume 
is  given.     One  for  finding  the  diameter  when  the  volume  is  given. 

4.  A  ball  of  diameter  2  in.  is  inside  a  right  circular  cylinder  of  diameter 
and  altitude  each  2  in.  How  many  cubic  inches  of  water  can  be  poured 
into  the  cylinder? 

5.  Find  the  weight  of  a  spherical  cast  iron  ball  of  diameter  6  in.,  if 
cast  iron  weighs  0.26  lb.  per  cubic  inch. 

6.  Find  the  volume  of  a  hollow  spherical  shell,  the  outer  diameter 
being  8  in.  and  the  thickness  2  in.  Ans.     234.6  cu.  in. 

7.  How  many  spherical  bullets  -f  of  an  inch  in  diameter  can  be 
made  from  5  pounds  of  lead,  if  lead  weighs  0.412  pounds  per  cubic  inch? 

8.  The  volume  of  two  spheres  are  in  the  ratio  of  512  :  27.  Find 
the  radius  of  each  if  the  sum  of  their  radii  is  44  in. 

9.  A  water  tank,  having  a  total  length  of  6  ft.  and  a  diameter  of  18 
in.,  is  in  the  form  of  a  right  circular  cylinder  with  two  hemispherical 
ends.     Find  its  capacity  in  gallons.  Ans.     72.7+  gal. 


VOLUME  OF   A  SPHERE 


403 


820.  Spherical  Sector.  If  a  sphere  is  generated  by  the 
revolution  of  a  semicircle,  the  solid  generated  by  a  sector 
whose  arc  is  a  part  of  this  semicircle  is  called  a  spherical 
sector. 

A  spherical  sector  generated  by  a  circular  sector  revolving 
about  one  of  its  bounding  radii  is  called  a  spherical  cone. 


The  zone  generated  by  the  arc  of  the  generating  sector  is 
called  the  base  of  the  spherical  sector. 

821.  Theorem.  The  volume  of  a  spherical  sector  is  equal  to 
one-third  the  product  of  the  area  of  Us  base,  and  its  radius. 

EXERCISES 

1.  A  right  circular  cone  having  a  vertex  angle  of  60°  has  its  vertex 
at  the  center  of  a  sphere  4  ft.  in  diameter.  What  is  the  volume  cut  from 
the  sphere  by  the  cone? 

2.  Find  the  volume  of  a  spherical  sector  cut  from  a  sphere  4  ft.  in 
diameter,  and  whose  base  is  a  zone  having  an  altitude  of  8  in. 

822.  Spherical  Segment.  The  portion  of  the  volume  of  a 
sphere  contained  between  two  parallel  planes  that  intersect 
the  sphere  or  are  tangent  to  it,  is  called  a 
spherical  segment. 

The   sections   of   the  sphere  made  by  the 
two    parallel    planes    are   the   bases   of   the  J 
spherical  segment. 

If  one  of  the  parallel  planes  is  tangent  to  the  sphere,  the 
segment  is  called  a  spherical  segment  of  one  base. 

The  distance  between  the  parallel  planes  is  the  altitude  of 
the  spherical  segment. 


404  SOLID  GEOMETRY 

823.  Theorem.  The  volume  of  a  spherical  segment  is  given 
by  the  formula  V =%Trh(rl2+r22)+^7rhz,  where  h  denotes  the  alti- 
tude, and  ri  and  r2  the  radii  of  the  bases  of  the  segment. 

Given  the  spherical  segment  with  altitude 
h,  and  bases  ADB  and  EHF  having  radii  n 
and  r2  respectively. 

To   prove    V  =  %irh(rl2+r22)+%7rh\ 

Proof.    Let  OG  =  x, 

V  =  volume  of  spherical  sector  formed 
by  revolving  OBF+ volume  of  cone  O-ADB 
-volume  of  cone  O-EHF.  Why? 

(1)  V  =  %irr2h+lwr12(h+x)  -%7rr22x.  Why? 
To  reduce  to  the  form  desired  it  is  necessary  to  eliminate  ? 

and  x,  and  this  requires  two  other  equations.     They  are: 

(2)  r2  =  rl2+(h+x)2,  and  (3)  r2  =  r22+x2.  Why? 

From  (2)  and  (3)  x  =  r_lzll^E^ 

2h 

Substituting  this  value  of  x  in  (3), 

9     h*+r1*+r2*+2r12h2+2r22h2-2r12r22 
rz  = . 

\h2 
Substituting  these  values  of  x  and  r2  in  (1)  and  simplifying, 

V  =  ±irh(n2+r22)+lTh\ 
Remark.    The  same  formula  is  derived  if  the  center  of  the 
sphere  is  within  the  spherical  segment. 

If  the  spherical  segment  has  but  one  base,  n  =  0,  and  the 
formula  becomes  V  =  ^irhr22+\irhz. 

The  student  should  carry  out  the  work  in  each  of  these  cases. 

EXERCISES 

1.  What  part  of  the  volume  of  the  earth  is  between  the  planes  cutting 
its  surface  in  the  parallels  of  30°  and  45°  north  latitude?  What  part 
is  north  of  a  plane  cutting  the  surface  in  the  parallel  of  60°  north  latitude? 

2.  In  the  formula  for  the  volume  of  a  spherical  segment,  solve  for  each 
letter  so  far  as  possible. 


VOLUME  OF  A  SPHERE 


405 


824.  Spherical  Wedge.  That  portion  of  the  volume  of  a 
sphere  contained  between  the  planes  of  two  great  semicircles 
is  called  a  spherical  wedge.  Its  curved  surface  is  its  base,  and 
is,  evidently,  a  lune. 


The  solid  A-NS-B  is  a  wedge.  The  lune  between  the  semicircles  NAS 
and  NBS  is  the  base  of  the  wedge.  The  angle  AOB  of  the  wedge  is  the 
angle  of  its  lune. 

825.  Theorem.  The  volume  of  a  spherical  wedge  is  equal  to 
one-third  the  product  of  the  area  of  its  base  and  its  radius.     That 

is,  V  = (f^rr3),  where  r  denotes  the  radius  of  the  sphere  and  u 

360 

the  angle  of  the  lune  in  degrees. 

By  a  discussion  similar  to  that  for  the  volume  of  a  sphere, 

the  volume  of  a  wedge  is  given  by  the  formula  V  =  \Lr,  where 

L  is  the  area  of  the  lune. 


u 


But  L  =  — (4  tit2)  by  §811. 
360 


y=- 


360 


(iirr*). 


EXERCISES 

1.  Find  the  volume  of  a  spherical  wedge  whose  angle  is  20°,  cut  from 
a  sphere  whose  radius  is  12  in. 

2.  Find  the  angle  of  a  spherical  wedge  if  its  volume  is  418.88  cu.  in., 
and  the  radius  of  the  sphere  is  10  in. 

3.  A  circular  flower  bed  is  in  the  form  of  a  spherical  segment  that  is 
25  ft.  in  diameter  and  2j  ft.  in  altitude.  How  many  loads  of  dirt  did  it 
take  to  build  it  up  if  one  load  is  \\  cu.  yd.? 

4.  The  figure  is  a  vertical  cross  section  of  a      V~ 16  /-tf'T 

casting,   the  inner  and   outer   "skins"   being  \  V  /    /   i 

spherical  zones.  Find  the  weight  of  metal  at  x^Ss"^*-— **£— . JL  1  ** 
0.35  lb.  per  cubic  inch  necessary  to  make  the  \^^  ^^/  _J 
casting.  Ans.     176  lb. 


406 


SOLID   GEOMETRY 


6.  Find  the  volume  of  the  segment  between  two  parallel  planes  6  in. 
apart  that  cut  a  sphere  having  a  radius  of  12  in.,  if  one  plane  passes  2 
in.  from  the  center.  There  are  two  cases:  (1)  when  the  center  of  the 
sphere  lies  outside  the  segment,  and  (2)  when  the  center  lies  in  the  seg- 
ment. Ans.     (1)  2186.6  cu.  in.;  (2)  2638.9  cu.  in. 

6.  The  number  of  square  inches  in  the  area  of  a  certain  sphere  is 
equal  to  four  times  the  number  of  cubic  inches  in  its  volume.  Find  the 
radius  of  the  sphere. 

7.  A  copper  ball  to  be  used  as  a  float  must  contain  22,449.4  cu.  in. 
What  must  be  its  diameter?  Ans.     35  in. 

8.  Derive  the  formula  for  the  volume  of  a  sphere  from  the  formula 
for  the  volume  of  a  spherical  sector. 

9.  Derive  the  formula  for  the  volume  of  a  sphere  from  the  formula 
for  the  volume  of  a  spherical  segment. 

Suggestion.     Let  n  =  0  and  r2=r  and  obtain  the  volume  of  a  hemisphere. 

10.  If  oranges  2 \  in.  in  diameter  are  selling  for  30  cents  a  dozen, 
what  should  be  the  price  for  oranges  3  in.  in  diameter? 

11.  The  water  tank  shown  in  the  figure  is  in  the 
form  of  a  right  circular  cylinder  with  a  hemisphere  at 
the  bottom.  Find  the  capacity  in  gallons  if  the 
diameter  is  22  ft.  and  the  altitude  of  the  cylinder  is 
24  ft.  6  in. 

12.  Find  how  many  square  feet  of  sheet  steel  in 
the  tank  if  it  is  open  at  the  top. 

13.  The  roof  over  the  tank  is  third  pitch  and  pro- 
jects 2  ft.  at  the  eaves.  Find  the  number  of  square 
feet  in  it. 

14.  A  shop  received  an  order  for  100  twelve-inch  steel  balls  to  be  used 
in  crushing  cement  in  a  cement  mill.  They  were  made  from  8^-inch 
round  steel  shafting.  What  length  of  shaft  was  taken  for  each  ball  if  5% 
was  allowed  for  scale? 

15.  A  round  or  button-head  machine  screw 
has  a  head  in  the  form  of  a  spherical  segment  of 
one  base.  Show  that  the  volume  of  the  head 
is  given  by  the  formula:  V —Trh{\w2-\-\h2)y 
where  w  is  the  diameter  of  the  head  and  h  its  height. 

16.  Find  the  volume  of  the  head  of  a  round  head  machine  screw  if  the 
diameter  of  the  head  is  0.731  in.  and  the  height  is  0.279  in. 

Ans.    0.070  cu.  in. 


SPHERICAL  ANGLES  407 


SPHERICAL  ANGLES 


826.  Angles  Formed  by  Arcs.  The  angle  formed  by  two 
arcs  of  circles  is  the  plane  angle  formed  by  the  lines  tangent 
to  the  arcs  at  the  point  of  intersection. 

Thus  the  angle  formed  by  the  arcs  AB  and  CB  is  the  #* 


angle  formed  by  the  tangent  lines  DB  and  EB.  "\   \c 

\A 

According  to  this  definition,  any  two  intersect- 
ing circles  on  a  sphere  form  angles.     In  this  text,  however,  only 
those  angles  formed  by  great  circles  are  considered. 

827.  Spherical  Angle.  The  angle  between  two  great  circle 
arcs  is  called  a  spherical  angle.  The  point  where  the  two 
arcs  meet  is  the  vertex  of  the  angle,  and  the  arcs  are  the 
sides  of  the  angle. 

The  terms  right,  acute,  and  obtuse  are  applied  to  spherical  angles  and 
have  the  same  meaning  as  with  plane  angles. 

828.  Since  the  planes  of  two  great  circles  intersect  in  a 
diameter  (§  767),  and  the  tangents  at  the  points  of  intersec- 
tion are  perpendicular  to  this  diameter  (§  288), 

the  angle  formed  by  the  tangents  is  the  plane 
angle  of  the  diedral  angle  formed  by  the  planes 
of  the  great  circles  §  588.  It  follows  then  from 
§826  that: 

A  spherical  angle  equals  in  degrees  the  diedral 
angle  between  the  planes  of  its  sides.  s 

Spherical  ZANB  =  ZCND  =  °diedral  ZA-NS-B. 

EXERCISES 

1.  Form  definitions  for  the  following  when  applied  to  spherical  angles: 
adjacent,  complementary,  supplementary,  vertical. 

2.  Prove  that  the  sum  of  all  the  spherical  angles  about  a  point  on  a 
sphere  is  equal  to  360°. 

3.  At  what  angle  does  a  meridian  of  the  earth  intersect  the  equator? 

4.  Prove  that  vertical  spherical  angles  are  equal. 


408  SOLID  GEOMETRY 

829.  Theorem.  A  sperical  angle  is  equal  in  degrees  to  the 
arc  of  the  great  circle  described  from  its  vertex  as  a  pole,  and 
included  between  its  sides,  produced  if  necessary.  P 

Given  spherical  ZAPB,  AB  an  arc  of  a  great  jt       J  |Y\ 

circle  whose  pole  is  P,  and  which  is  included  L ■*-  —  AA 

between  the  sides  AP  and  BP.  \            \ni 

To  prove  that  spherical  ZAPB  =  °AB.  \"         J 

Proof.    Draw  PO,  AO,  and  BO.  ^""1^ 

PO  is  perpendicular  to  plane  of  AB.  §  775 

Then  OA  _L  PO  and  BO  JL  PO.  Why? 
And  ZAO£  is  the  plane  angle  of  diedral  ZA-QP-B.       Why? 

Hence  diedral  ZA-OP-B  =  °ZAOB.  §  589  (3) 

But                             ZAOB  =  °AB.  §  305 

.-.  spherical  ZAPB  =  °AB.  §  111 

SPHERICAL   POLYGONS  AND   POLYEDRAL  ANGLES 

830.  Spherical  Polygons.  A  portion  of  a  sphere  bounded 
by  three  or  more  arcs  of  great  circles  is  called  a  spherical 
polygon.  The  bounding  arcs  are  the  sides 
of  the  polygon,  the  spherical  angles  formed 
by  the  sides  are  the  angles  of  the  polygon, 
and  their  vertices  the  vertices  of  the  poly- 
gon. 

831.  A  diagonal  of  a  spherical  polygon  is 
the  arc  of  a  great  circle  joining  any  two  vertices  not  adjacent. 

It  is  evident  that  a  side  of  a  spherical  polygon  can  be  greater  than  a 
semicircle,  but  in  this  text  only  those  will  be  considered  whose  sides  are 
each  less  than  a  semicircle. 

832.  A  spherical  triangle  is  a  spherical  polygon  of  three 
sides. 

The  terms  right,  acute,  obtuse,  isosceles,  and  equilateral  are  applied  to 
spherical  triangles  and  have  the  same  meaning  as  with  plane  triangles. 
It  will  be  found,  however,  that  a  spherical  triangle  may  have  one,  two, 
or  three  right  or  obtuse  angles. 


SPHERICAL  POLYGONS  AND  POLYEDRAL  ANGLES     409 

833.  Central  Polyedral  Angles.  It  follows  from  the  defini- 
tion of  a  spherical  polygon  that  the  planes  of  the  sides  of  the 
polygon  pass  through  the  center  of  the 
sphere.  These  planes  form  a  polyedral 
angle,  called  the  corresponding  central 
polyedral  angle  of  the  spherical  polygon. 
This  polyedral  angle  and  the  spherical 
polygon  are  so  related  that  it  is  convenient 
to  consider  them  in  connection  with  each 
other. 

It  follows  directly  from  §§  305  and  829  that: 

(1)  The  sides  of  a  spherical  polygon  are  equal  in  degrees  to  the 
face  angles  of  the  polyedral  angle. 

(2)  The  angles  of  a  spherical  polygon  are  equal  in  degrees  to 
the  diedral  angles  of  the  polyedral  angle. 

Because  of  these  relations  it  follows  that  to  each  property  of 
a  polyedral  angle  there  is  a  corresponding  property  of  the  spher- 
ical polygon,  and  vice  versa.  One  of  these  properties  can  be 
stated  when  the  other  is  given  by  making  the  substitution 
indicated  in  the  following: 


Spherical  Polygon 
Sides 

Angles  of  polygon 
Vertices  of  polygon 
Center  of  sphere 


Polyedral  Angle 
Face  angles 
Diedral  angles 
Edges 
Vertex 


In  this  manner  a  theorem  relating  to  spherical  polygons  can 
readily  be  worded  so  as  to  apply  to  polyedral  angles,  and  vice 


834.  A  convex  spherical  polygon  is  a  spherical  polygon  no 
pide  of  which,  when  produced,  will  enter  the  polygon. 

It  is  evident  that  the  corresponding  central  polyedral  angle 
of  a  convex  spherical  polygon  is  a  convex  polyedral  angle. 
(See  §  605.) 


410 


SOLID  GEOMETRY 


835.  Two  spherical  polygons  on  the  same  sphere  or  equal 
spheres  are  congruent  if  their  corresponding  central  polyedral 
angles  are  congruent.     (See  §  607.) 

The  spheres  0  and  0'  are  equal  and 
the  polyedral  angles  having  their  ver- 
tices at  the  centers  of  the  spheres  are 
congruent.  Then  the  corresponding 
spherical  polygons  ABCD  and  A'B'C'D'  are  congruent. 

836.  Symmetric  Polyedral  Angles  and  Spherical  Polygons. 

Two  polyedral  angles  are  symmetric  if  the  face  angles  and 
diedral  angles  of  one  are  respectively  equal  to  those  of  the 
other  but  arranged  in  reverse  order. 

The  triedral  angles  V-ABC  and  V'-A'B'C  are  symmetric,  for  diedral 
angles  AV,  BVf  and  CV  are  equal  respec- 
tively to  diedral  angles  A'V,  B'V,  and 
C'F'j  and  the  face  angles  AVB,  BVC,  and 
CVA  are  equal  respectively  to  face  angles 
A'V'B,  B'V'C,  and  C'V'A'. 

It  is  to  be  noticed  that  if  the  triedral 
angles  are  viewed  from  the  vertices  V  and 
V,   then   the  parts  of  the  triedral  angle   V-ABC  are  arranged  in  the 
reverse  order  to  the  parts  of  V'-A'B'C'. 

If  the  two  triedral  angles  were  made  of  pieces  of  cloth  sewed  together, 
then,  if  one  were  turned  "inside  out,"  they  could  be  made  to  coincide. 
They  would  then  be  congruent  polyedral  angles. 

A  pair  of  gloves  are  analogous  to  two  symmetric  polyedral  angles. 
All  parts  of  one  are  equal  to  corresponding  parts  of  the  other,  but  are 
arranged  in  reverse  order.  Right  and  left  parts  of  animals  are  symmetric 
to  each  other  in  this  sense.  Numerous  such  illustrations  of  symmetry  can 
be  found  in  nature  and  in  the  arts. 

837.  Two  spherical  polygons  on  the  same  sphere  or  an 
equal   spheres   are   symmetric  if   their    corre- 
sponding central  polyedral  angles  are  sym- 
metric. 

In  the  figure,  spherical  triangles  ABC  and  A'B'C 
are  symmetric  for  their  corresponding  central  triedral 
angles  are  symmetric. 


SPHERICAL  POLYGONS  AND  POLYEDRAL  ANGLES    411 


838.  The  two  polyedral  angles  formed  by  a  pyramidal  surface 
are  symmetric;  and,  if  a  sphere  is  drawn  with  its  center  at  the 
vertex  of  the  pyramidal  surface,  then  the  pyramidal  surface  will 
intersect  the  sphere  in  two  symmetric  spherical 
polygons. 

It  is  evident  that  the  parts  of  the  two  nappes  of  the 
pyramidal  surface  with  vertex  0  are  arranged  in  reverse  j 
order,  then  the  corresponding  spherical  polygons  ABCD  \ 
and  A'B'C'D'  are  symmetric. 

Of  course,  symmetric  spherical  polygons  can  He  in 
any  position  on  the  sphere,  for  a  figure  on  a  sphere  can 
be  moved  about  without  changing  its  shape. 

839.  Theorem.  In  any  spherical  triangle,  the 
sum  of  two  sides  is  greater  than  the  third  side. 

Given  spherical  AABC  on  sphere  0. 
To  prove  that  AB+BC>AC. 
Proof.    ZAOB+ZBOOZAOC.  §608 

But  ZAOB  =  °AB,  ZBOC  =  °BC,  and  ZAOC  =  °AC. 
.\AB+BC>AC. 

840.  Theorem.  In  any  convex  spherical  poly- 
gon, the  sum  of  the  sides  is  less  than  a  great  circle  of 
the  sphere,  or  less  than  360°. 

ZAOB+ZBOC+ZCOD+ZDOA<3W°.    § 609 
.*.  AB+BC+CD+DA<Sm°.        Why? 


EXERCISES 

1.  Prove  that  any  side  of  a  spherical  polygon  is  less  than  the  sum  of 
the  remaining  sides. 

2.  Show  that  any  side  of  a  convex  spherical  polygon  is  less  than  180°. 

3.  Given  a  spherical  AABC,  with  AB  =  57°  and  AC  =  125°.  Between 
what  limits  must  BC  lie? 

4.  Given  a  spherical  quadrilateral  three  of  whose  sides  are  77°,  98°, 
and  45°  respectively.  Between  what  limits  must  the  fourth  side  he? 
Between  what  limits  must  the  fourth  side  he  if  three  sides  are  27°,  33°, 
and  100°  respectively? 


412  SOLID  GEOMETRY 

841.  Theorem.  Two  triedral  angles  are  congruent  or  sym- 
metric if  a  face  angle  and  the  two  adjacent  diedral  angles  oj  one  are 
equal  respectively  to  a  face  angle  and  the  two  adjacent  diedral 
angles  of  the  other. 

Given  triedral  AV-ABC  and  V'-DEF, 
with  ZA  VC  =  ZDV'F,  and  the  adjacent 
diedral  angles  equal. 

To  prove  that  the  triedral  angles  are 
congruent  or  symmetric. 

Proof.  If  parts  are  arranged  in  the 
same  order,  superpose  as  in  the  analogous  case  in  plane  geometry 
(§  248).  If  parts  are  not  in  the  same  order,  extend  the  edges 
through  the  vertex  of  one  and  superpose  the  other  upon  the 
symmetric  triedral  angle  thus  formed.  §  836 

842.  Theorem.  On  the  same  sphere  or  on  equal  spheres ,  two 
spherical  triangles  are  congruent  or  symmetric  if  a  side  and  the 
two  adjacent  angles  of  one  are  equal  respectively  to  a  side  and  the 
two  adjacent  angles  of  the  other. 

Construct  corresponding  central  triedral  angles  and  apply  the  previous 
theorem. 

843.  Theorem.  Two  isosceles  symmetric  spherical  triangles 
are  congruent. 

844.  Theorem.  Two  triedral  angles  are  congruent  or  sym- 
metric if  two  face  angles  and  the  included  diedral  angle  of  one 
equal  respectively  to  two  face  angles  and  the  included  diedral  angle 
of  the  other. 

Superpose  if  arranged  in  the  same  order.  If  arranged  in  opposite 
order,  superpose  one  upon  the  triedral  angle  symmetric  to  the  other. 

845.  Theorem.  On  the  same  sphere  or  on  equal  spheres,  two 
spherical  triangles  are  congruent  or  symmetric  if  two  sides  and  the 
included  angle  of  one  are  equal  respectively  to  two  sides  and  the 
included  angle  of  the  other. 

Construct  corresponding  central  triedral  angles  and  apply  the  previous 
theorem. 


SPHERICAL  POLYGONS  AND  POLYEDRAL  ANGLES    413 

846.  Theorem.  On  the  same  sphere  or  on  equal  spheres^  two 
spherical  triangles  are  congruent  or  symmetric  if  the  three  sides  of 
one  are  equal  respectively  to  the  three  sides  of  the  other. 

Construct  the  corresponding  central  triedral  angles  and  use  §  610  when 
parts  are  in  the  same  order.  When  parts  are  in  the  opposite  order,  con- 
struct a  triedral  angle  symmetric  to  one  by  §  836,  and  then  apply  §  610. 

847.  A  triedral  angle  is  isosceles  if  it  has  two  equal  face 
angles.    It  is  equilateral  if  it  has  all  its  face  angles  equal. 

848.  Theorem.  If  a  spherical  triangle  is  isosceles,  the  angles 
opposite  the  equal  sides  are  equal. 

Given    the    isosceles    spherical    AABC,    with 
AB  =  CB. 
To  prove  ZA  =  ZC. 

Proof.    Draw  the  great  circle  arc  BD  bisecting 
AC. 

Then  spherical  A  ABD  and  CBD  are  congruent  or  symmetric. 

§846 
.-.  ZA  =  ZC.  Why? 

EXERCISES 

1.  If  a  triedral  angle  is  isosceles,  the  diedral  angles  opposite  the  equal 
face  angles  are  equal. 

2.  If  a  spherical  triangle  is  equilateral,  it  is  equiangular. 

3.  If  a  triedral  angle  is  equilateral,  all  the  diedral  angles  are  equal. 

4.  State  and  prove  theorems  on  the  sphere  corresponding  to  the  fol- 
lowing theorems  in  plane  geometry: 

(1)  All  points  in  the  perpendicular  bisector  of  a  straight  line  are  equi- 
distant from  the  extremities  of  the  line.  §  205 

(2)  All  points  equidistant  from  the  extremities  of  a  straight  line  lie  in 
the  perpendicular  bisector  of  the  line.  §  206 

(3)  All  points  in  the  bisector  of  an  angle  are  equidistant  from  the  sides 
of  the  angle.  §  199 

5.  In  the  isosceles  triangle  of  §  848,  could  AB  and  CB  each  be  greater 
than  90°?  Could  each  be  greater  than  180°?  Could  ZABC  be  20°  and 
AC  be  5°?     Could  ZABC  be  20°  and  AC  be  20°?     Illustrate  on  a  globe. 


414  SOLID  GEOMETRY 

849.  A  triedral  angle  is  called  a  rectangular,  a  birectangu- 
lar,  or  a  trirectangular  triedral  angle,  according  as  it  has  one, 
two,  or  three  right  diedral  angles. 

Three  concurrent  faces  of  a  rectangular  solid  form  a  trirectangular 
triedral  angle.  Three  concurrent  faces  of  a  right  prism  with  an  oblique 
triangular  base  form  a  birect angular  triedral  angle. 

850.  A  spherical  triangle  is  called  a  right,  a  birectangular, 
or  a  trirectangular  spherical  triangle,  according  as  it  ha£  one, 
two,  or  three  right  spherical  angles. 

851.  The  following  are  derived  from  §§779  and  785: 

(1)  In  a  birectangular  spherical  triangle,  the  sides  opposite  the 
equal  angles  are  quadrants,  and  conversely. 


In  the  figure,  0  is  a  birectangular  triedral  angle,  and  APB  is  a  birec- 
tangular spherical  triangle.  The  faces,  or  arcs,  that  pass  through  P  form 
right  angles  with  the  other  face,  or  arc. 

(2)  In  a  trirectangular  spherical  triangle,  the  sides  are  quad- 
rants, and  conversely. 


In  the  figure,  0  is  a  trirectangular  triedral  angle,  and  APB  is  a  trirec- 
tangular spherical  triangle.  Here  each  vertex  is  the  pole  of  the  opposite 
side  of  the  spherical  triangle.  §  785 

852.  Theorem.  Three  mutually' perpendicular  planes  passed 
through  the  center  of  a  sphere  divide  the  sphere  into  eight  con- 
gruent trirectangular  spherical  triangles. 


POLAR  TRIANGLES  415 

EXERCISES 

1.  Find  the  area  of  a  trirectangular  spherical  triangle  on  a  sphere 
having  a  radius  of  10  in. 

2.  Find  the  area  of  a  birectangular  spherical  triangle  with  a  vertex 
angle  of  60°,  on  a  sphere  40  ft.  in  diameter. 

3.  What  are  the  volumes  cut  out  of  the  sphere  by  the  corresponding 
central  triedral  angles  in  Exercises  1  and  2? 

4.  On  a  sphere  having  a  radius  of  12  in.  two  sides  of  an  isosceles  spher- 
ical triangle  are  quadrants.  Find  the  length  of  the  third  side  if  the  vertex 
angle  is  60°. 

5.  On  a  sphere  having  a  radius  of  4  in.  the  sides  of  a  spherical  triangle 
are  80°,  140°,  and  120°  respectively.     Find  the  length  of  each  side. 

6.  Find  the  area  of  a  birectangular  spherical  triangle  having  one  angle 
1°,  on  a  sphere  of  radius  r. 

7.  What  is  the  volume  cut  out  of  the  sphere  by  the  central  triedral 
angle  corresponding  to  the  spherical  triangle  described  in  Exercise  6? 

POLAR  TRIANGLES 

853.  If  from  the  vertices  of  the  spherical  triangle  ABC  as 
poles,  great  circles  are  drawn,  the  sphere  is  divided  into  eight 
spherical  triangles.  If  the  intersection  of  the 
circles  having  A  and  B  as  poles,  that  is  on  the 
same  side  of  AB  as  C,  is  labeled  C,  and  simi- 
larly for  the  vertices  A'  and  B';  then  the  spher- 
ical triangle  A'B'C  is  the  polar  triangle  of 
ABC. 

"On  the  same  side  of"  means  that  C  and  C"  are  both  in  one  of  the 
hemispheres  formed  by  the  great  circle  of  which  AB  is  an  arc. 

Two  spherical  triangles  related  as  ABC  and  A'B'C  are  spoken 
of  as  polar  triangles. 

Other  spherical  triangles  should  be  constructed,  some  having 
sides  of  few  degrees  and  others  having  sides  near  180°.  The 
polar  triangles  of  these  should  be  drawn  and  labeled  as  directed. 

It  is  important  that  these  relations  should  be  clearly  under- 
stood, as  they  are  used  in  proving  many  theorems. 


416 


SOLID   GEOMETRY 


854.    Theorem.     //  one  spherical   triangle  is  the  polar  of 
another,  then  the  second  is  the  polar  of  the  first. 

Given  the  spherical  AABC  and  its  polar 
AA'B'C. 

To  prove  that  AABC  is  the  polar  triangle  of 
AA'B'C'. 

Proof.    A  is  the  pole  of  jfC',  and  C  is  the 
pole  of  A^B'.  Given 

Then  B  is  at  a  quadrant's  distance  from  both  A  and  C.  Why? 

Hence  Bf  is  the  pole  of  AC.  §  787 

Similarly  A1  is  the  pole  of  BC,  and  C  the  pole  of  AB. 

.'.  AABC  is  the  polar  triangle  of  AA'B'C.  §  853 


EXERCISES 

1.  One  side  of  a  spherical  triangle  on  the  earth's  surface  is  on  the 
equator.  Prove  that  a  vertex  of  the  polar  triangle  is  either  at  the  north 
pole  or  at  the  south  pole  of  the  earth. 

2.  Prove  the  theorem  of  §  854  by  taking  as  given  that  AABC  is  the 
polar  triangle  of  AA'B'C. 

3.  If  a  vertex  of  a  spherical  triangle  is  at  the  pole  of  the  great  circle  of 
which  one  side  is  an  arc,  prove  that  one  vertex  of  the  polar  triangle  is  at 
this  pole  and  that  one  side  is  on  this  great  circle. 

4.  Draw  the  following  spherical  triangles  and  their  polar  triangle 

(1)  A  triangle  each  side  of  which  is  less  than  a  quadrant. 

(2)  A  triangle  each  side  of  which  is  greater  than  a  quadrant. 

(3)  A  triangle  only  one  side  of  which  is  greater  than  a  quadrant. 

(4)  A  birectangular  triangle. 

(5)  A  trirectangular  triangle. 

Note.  A  slated  sphere  should  be  used  in  making 
these  constructions.  A  pair  of  compasses  should  be 
so  adjusted  that  they  will  always  strike  an  arc  of  a 
great  circle  on  the  sphere  being  used. 

5.  Can  the  sides  of  a  spherical  triangle  intersect 
the  sides  of  its  polar  triangle?    Under  what  conditions? 

6.  Prove  the  theorem  of  §  854  by  using  this  figure. 

7.  Prove  that  a  trirectangular  triangle  is  its  own  polar  triangle. 


POLAR  TRIANGLES  417 

855.  Theorem.  In  two  polar  triangles,  each  angle  of  the 
one  is  equal  in  degrees  to  the  supplement  of  the  side  opposite 
to  it  in  the  other. 


Given  the  polar  triangles  ABC  and  A'B'C ',  with  the  letter 
at  each  vertex  of  an  angle  denoting  its  value  in  degrees,  and  the 
letters  a,  b,  c  and  a',  bf,  c'  denoting  the  values  of  the  opposite 
sides  in  degrees. 

To  prove  that  A  +a'  =  180°,  B  +6' =  180°,  C+  c'  =  180°; 
and  A'+a  =180°,  B'+b  =180°,  C'+c  =180°. 

Proof.  Let  D  and  E  respectively  be  the  intersections  of  A  B 
and  AC,  produced  if  necessary,  with  B'C'. 

Then  CD  =  90°,  and  KB'  =  90°.  §  785 

Or  CE+ED  =  90°,  and  ED+DB'  =  90°.  Why? 

Then  CE '+ ED + ED + DB'  =  180°.  Why? 

But  C^+ED+DB'  =  a',  and  ED  =  °ZA.      §§109,829 

.-.A  +a/  =  180°.  Why? 

Similarly  B+b'  =  180°,  and  C+c'  =  180°. 

Complete  the  proof  by  starting  with  AA'B'C. 

EXERCISES 

1.  If  the  sides  of  a  spherical  triangle  have  respectively  70°,  80°,  and 
110°,  find  the  values  of  the  angles  of  its  polar  triangle. 

2.  Could  the  sides  of  a  spherical  triangle  be  each  8°?  What  would 
then  be  the  value  of  the  angles  in  the  polar  triangle? 

3.  Could  the  angles  of  a  spherical  triangle  be  8°,  9°,  and  10°  respectively? 
What  would  then  be  the  sides  of  the  polar  triangle?   . 

4.  Find  the  area  of  the  triangle  that  is  the  polar  of  a  bireetangular 
spherical  triangle  having  an  angle  of  1°,  on  a  sphere  4  ft,  in  radius. 


418 


SOLID   GEOMETRY 


856.  Theorem.  If  two  spherical  triangles  on  the  same  sphere 
or  on  equal  spheres  are  mutually  equiangular,  they  are  mutually 
equilateral,  and  are  either  congruent  or  symmetric. 


Given  two  mutually  equiangular  spherical  triangles  Q  and 
R  on  equal  spheres. 

To  prove  that  Q  and  R  are  mutually  equilateral,  and  are 
either  congruent  or  symmetric. 

Proof.    Construct  A  Qf  and  R',  the  polar  triangles  of  A  Q  and 
R  respectively. 

A  Q  and  R  are  mutually  equiangular. 

Then  A  Qf  and  Rf  are  mutually  equilateral. 

Hence  A  Qf  and  Rr  are  either  congruent  or  symmetric. 

And  A  Q'  and  Rf  are  mutually  equiangular. 

But  A  Q  and  R  are  the  polar  triangles  of  A  Qf  and  Rf. 

Then  A  Q  and  R  are  mutually  equilateral. 

.%  A  Q  and  R  are  either  congruent  or  symmetric. 


Given 
§855 
§846 

Why? 

§854 
§855 
§846 


857.  Theorem.  If  two  triedral  angles  have  their  diedral 
a?igles  respectively  equal,  their  face  angles  are  respectively  equal, 
and  they  are  either  congruent  or  symmetric. 

EXERCISES 

1.  Is  there  a  theorem  in  plane  geometry  analogous  to  the  theorems 
of  §§  856,  857? 

2.  If  two  angles  of  a  spherical  triangle  are  equal,  the  triangle  is  isosceles. 

3.  If  a  spherical  triangle  is  equiangular  it  is  equilateral. 

4.  If  two  angles  of  a  spherical  triangle  are  unequal,  the  sides  opposite 
these  angles  are  unequal,  and  the  greater  side  is  opposite  the  greater  angle; 
and  conversely,    Proof  of  theorem  is  similar  to  that  of  §  184. 


POLAR  TRIANGLES  419 

858.    Theorem.     The  sum  of  the  angles  of  a  spherical  triangle 
is  greater  than  180°  and  less  than  540°. 


Given  the  spherical  AABC  with  the  letter  at  each  vertex  of 
an  angle  denoting  its  value  in  degrees,  and  the  letters  a,  b,  c 
denoting  the  values  of  the  opposite  sides  in  degrees. 

To  prove  A+B +0180°  and  <540°. 

Proof.    Construct  the  polar  AA'B'C  of  AABC. 

Then        A +a'  =  180°,  B+&'=180°,  C+c'  =  180°.  §855 

Hence  A+B+C+a'+b'+cf  =  540°.  §  105 

But  a'+6'+c'<360°.  §840 

.\A+£+C>180°.  §  178 

Also  a'+bf+cf>0°.  Why? 

.\A+B+C<540°.  Why? 

859.  Theorem.  The  sum  of  the  diedral  angles  of  a  triedral 
angle  is  greater  than  180°  and  less  than  51fi>°. 

EXERCISES 

1.  How  would  you  draw  a  triangle  on  a  sphere,  the  sum  of  the  angles 
of  which  would  be  only  a  little  greater  than  180°?  Only  a  little  less  than 
540°?  Could  you  draw  such  triangles  on  a  small  sphere  or  would  it  be 
necessary  to  have  a  large  sphere? 

2.  What  is  the  locus  of  points  on  a  sphere  and  equally  distant  from  two 
intersecting  arcs  of  great  circles  of  the  sphere? 

3.  What  is  the  locus  of  the  points  on  a  sphere  that  are  equally  distant 
from  the  ends  of  an  arc  of  a  great  circle? 

4.  If  the  sides  of  a  spherical  triangle  each  have  less  than  3°,  what 
can  be  said  of  the  angles  of  its  polar  triangle?  Could  the  angles  of  a  spher- 
ical triangle  be  respectively  170°,  (S°,  and  12°? 


420  SOLID  GEOMETRY 

AREAS   OF   SPHERICAL   POLYGONS 

860.    Theorem.     On  the  same  sphere  or  on  equal  spheres  two 
symmetric  spherical  triangles  are  equal  in  area. 


Given  the  two  symmetric  spherical  triangles  ABC  &nd  A' B'C. 

To  prove  that  AABC  =  AA'B'C 

Proof.    Place  the  two  triangles  so  that  they  are  formed  by 
the  same  pyramidal  surface  with  its  vertex  at  0.  §  838 

Let  P  be  the  pole  of  a  small  circle  passing  through  A,  B, 
and  C. 

Draw  the  diameter  POP'  and  the  great  circle  arcs  PA,  PB, 
PC,  P'A',  P'B',  and  P'C. 

Then  PA=PB  =  PC.  §783 

But  P^A^PA,  P'B'^PB,  fiV^PC.  Why? 

Then  P7A,  =  i%,  =  P7C^  Why? 

And  APAB  and  P'A'B'  are  isosceles  as  well  as  symmetric. 

Hence  APAB^AP'A'B'.  §843 

Similarly   APBC^AP'B'C,  and  APCA^AP'C'A'. 

Then         APAB+ APBC+ APCA  =  AP'A'B'+ 

AP'B'C'+AP'C'A'.  §  105 

.-.  AABC  =  AA,B,C.  §§  109,  104 

EXERCISES 

1.  Draw  a  figure  in  which  the  pole  P,  §  860,  falls  outside  the  triangle 
ABC.    What  difference  would  this  make  in  the  proof? 

2.  A  sphere  can  be  divided  into  four,  eight,  or  twenty  equal  equilateral 
spherical  triangles. 


AREAS  OF  SPHERICAL  POLYGONS  421 

861.  Spherical  Excess.  The  excess  of  the  sum  of  the 
angles  of  a  spherical  triangle  over  180°  is  called  the  spherical 
excess  of  the  triangle. 

The  spherical  excess  of  a  spherical  polygon  of  n  sides  is  the 
excess  of  the  sum  of  its  angles  over  (n  —  2)180°,  that  is,  it  is 
the  excess  over  the  sum  of  the  angles  of  a  plane  polygon  of 
the  same  number  of  sides. 

862.  Area  of  a  Spherical  Triangle.  The  sides  as  well  as 
the  angles  of  a  spherical  triangle  are  usually  given  in  degrees. 
It  is  evident  then  that  the  number  of  square  units  in  the 
area  of  a  spherical  triangle  will  depend  upon  the  radius  of 
the  sphere  on  which  the  triangle  is.  When  the  radius  of  the 
sphere  is  not  given  it  is  convenient  to  have  a  unit  of  meas- 
ure that  is  some  fractional  part  of  the  surface  of  a  sphere. 

863.  Spherical  Degree.  One  half  the  ar,ea  of  a  lune  of  1°, 
or  the  area  of  a  birect angular  triangle  having  a  vertex  angle  of  1°, 
or  y^-Q  of  the  surface  of  a  sphere,  is  called  a  spherical  degree. 
It  is  the  unit  of  measure  of  spherical  polygons. 

A  polyedral  angle  whose  vertex  is  at  the  center  of  a 
sphere  cuts  a  spherical  polygon  out  of  the  sphere.  The 
area  of  this  polygon  in  spherical  degrees  is  called  the 
measure  of  the  polyedral  angle.  In  this  manner  the 
magnitudes  of  polyedral  angles  may  be  compared. 

EXERCISES 

1.  How  many  spherical  degrees  in  a  lune  of  20°?     Of  46°  30'? 

2.  How  many  spherical  degrees  in  a  trirectangular  spherical  triangle? 
What  is  the  area  in  square  inches  of  a  trirectangular  triangle  on  a  sphere 
having  a  radius  of   10  in.? 

3.  How  many  square  feet  in  the  area  of  a  birectangular  triangle  with 
vertex  angle  18°  on  a  sphere  6  ft.  in.  diameter? 

4.  Find  the  spherical  excess  in  spherical  triangles  having  the  following 
angles:  (1)  100°,  117°,  136°.  (2)  141°,  97°  30',  67°  45'.  (3)  116°  23'  14", 
127°  43'  18",  69°  45'  15". 

6.  Find  the  spherical  excess  in  a  spherical  pentagon  having  angles  of 
113°,  97°,  119°  17',  141°  23',  and  172°  27', 


422  SOLID   GEOMETRY 

864.  Theorem.  A  spherical  triangle  is  equivalent  to  a  lune  on 
the  same  sphere,  whose  angle  is  half  the  spherical  excess  of  the  tri- 
angle. 


Given  the  spherical  AABC  on  a  sphere  of  surface  S  and 
center  0,  with  A,  B,  and  C  denoting  the  value  in  degrees  of 
the  angles  at  the  respective  vertices. 

To  prove  that  AABC  =  a  lune  of  angle  ^(A+B+C -180°). 

Proof.    Complete  the  great  circles  of  the  sides  of  AABC,  and 

let  one  of  them  as  BC  from  the  boundary  of  the  hemisphere  which 

is  divided  into  four  triangles,  ABC,  or  X,  ABC  or  Y,  AB'C  or 

Z,  and  AB'C  or  U. 

ABA'C  on  the  reverse  hemisphere  =  AZ.  §  860 

Lune  of  ZA  =  AX+AZ.  Why? 

Lune  of  ZB  =  AX+ A*7.  Why? 

Lune  of  AC  =  AX+AY.  Why? 

Then     lune  of  Z(A+B+C)  =  ZAX+AZ+AU+AY.     §  105 

But      AX+AZ+AU+AY  =  ±S  =  lune  of  Z180°.  Why? 

Then     lune  of  Z(A+B+C)  =2AX+lune  of  Z180°.        §  111 

Or  2AX  =  lune  of  Z(A+B+C -180°).  Why? 

.-.  AABC  =  lune  of  Z\(A+B+C -180°).        Why? 

865.  Theorem.  On  the  same  sphere  or  on  equal  spheres,  the 
areas  of  two  spherical  triangles  are  to  each  other  as  their  spherical 
excesses. 

866.  Theorem.     //  A  denotes  the  area  of  a  spherical  triangle, 

E  the  number  of  degrees  in  its  spherical  excess,  and  r  the  radius  of 

irr2E 
the  sphere,  then  the  area  is  given  by  the  formula:   A  =-7^r« 


? 


AREAS  OF  SPHERICAL  POLYGONS        423 

867.  Theorem.  A  spherical  polygon  is  equivalent  to  a  lune  on 
the  same  sphere,  whose  angle  is  half  the  spherical  excess  of  the 
polygon. 

Divide  the  spherical  polygon  into  spherical  triangles  by  drawing  the 
diagonals  from  one  vertex.  Then  the  area  of  the  polygon  is  equivalent 
to  the  sum  of  the  areas  of  the  triangles,  and  the  spherical  excess  of  the 
polygon  is  equal  to  the  sum  of  the  excesses  of  the  triangles. 

868.  Theorem.     The  area  of  a  spherical  polygon  is  given  by 

7rr2E 
the  formula  A  — ,  where  r  is  the  radius  of  the  sphere  and  E  the 

number  of  degrees  in  the  spherical  excess. 

Note.  This  theorem  was  stated  by  A.  Girard  in  1629  and  was  first 
completely  proved  by  Cavalieri  a  few  years  later. 

869.  Spherical  Pyramid.     The  portion  of  the  volume  of  a 
sphere  bounded  by  a  spherical  polygon  and 
the  planes  of  its  sides  is   called   a  spherical 
pyramid. 

The  polygon  is  the  base  of  the  spherical 
pyramid,  and  the  center  of  the  sphere  is  its 
vertex. 

Thus,  O-ABCD  is  a  spherical  pyramid. 

870.  Theorem.  The  volume  of  a  spherical  pyramid  is  equal 
to  one-third  the  product  of  the  area  of  its  base  by  the  radius  of  the 
sphere. 

EXERCISES 

1.  Find  the  areas  of  triangles  on  spheres  of  the  given  radii,  and  having 
angles  as  follows: 

(1)  80°,  140°,  120°,  r  =  10  in.  (3)   125°,  135°,  145°,  r  =  50  ft. 

(2)  150°,  75°,  110°,  r  =  16  in.  (4)   179°,  179°,  179°,  r  =  10  ft. 

(5)  112°  30'  15",  127°  43'  30",  175°  27'  15",  r  =  20  ft. 

(6)  37°  43'  27",  15°  45'  34",  170°  45'  43",  r  =  4000  mi. 

2.  Find  the  areas  of  polygons  on  spheres  of  the  given  radii,  and  having 
angles  as  follows: 

(1)  70°,  170°,  100°,  125°,  r  =  12in. 

(2)  100°,  98°,  175°,  160°,  128°,  96°,  r  =  8  ft. 

(3)  75°  45'  17",  127°  14'  16",  102°  43'  16",  150°  43'  41",  r  =  20  in. 


424  SOLID  GEOMETRY 

3.  Find  the  area  of  the  triangle  on  the  earth's  surface,  determined  by 
the  meridians  at  48°  and  75°  west  longitude,  and  the  equator.  Use  4000 
miles  for  the  radius  of  the  earth. 

4.  Find  the  volume  of  a  spherical  pyramid,  given  that  its  base  is  an 
equiangular  triangle  having  each  angle  120°,  on  a  sphere  of  radius  9  in. 

5.  Find  the  volume  of  a  spherical  pyramid,  given  that  its  base  is  a 
spherical  hexagon  each  angle  of  which  is  130°,  on  a  sphere  of  12  in.  radius. 

6.  The  volume  of  a  spherical  pyramid  whose  base  is  an  equiangular 
spherical  triangle  with  angles  of  105°  is  2567T  cu.  in.  Find  the  diameter 
of  the  sphere. 

7.  What  is  the  greatest  area  that  a  triangle  on  the  surface  of  the  earth 
may  have  if  the  sum  of  its  angles  is  not  to  differ  by  more  than  1°  from 
180°?     If  the  sum  is  not  to  differ  by  more  than  1'  from  180°? 

8.  A  triangle  having  an  area  of  62  sq.  in.  is  on  a  sphere  that  has  an 
area  of  500  sq.  in.  Two  of  its  angles  are  92°  and  135°  respectively.  Find 
the  third  angle. 

9.  The  sides  of  a  spherical  triangle  are  79°,  88°,  and  115°,  and  it  is 
on  a  sphere  whose  radius  is  18  in.     Find  the  area  of  its  polar  triangle. 

QUESTIONS 

1.  Define  a  circle.  A  sphere.  Is  a  circle  an  area?  Is  a  sphere  a 
solid?     Define  a  sphere  as  a  locus.     Define  a  circle  as  a  locus. 

2.  Is  it  in  agreement  with  the  definition  of  a  sphere  to  speak  of  the 
area  of  a  sphere?  What  is  the  volume  of  a  sphere?  Compare  the  defini- 
tion of  the  area  of  a  circle  with  that  of  the  volume  of  a  sphere. 

3.  What  is  a  spherical  angle?  A  spherical  triangle?  Is  a  triangle 
an  area? 

4.  How  is  a  plane  angle  measured?  A  spherical  angle?  A  diedral 
angle?     A  polyedral  angle? 

5.  What  is  a  lune?  A  wedge?  How  may  lunes  be  compared  in 
size?     How  may  wedges  be  compared  in  size? 

6.  Define  the  sector  of  a  sphere,  and  compare  with  the  sector  of  a 
circle. 

7.  Define  the  segment  of  a  sphere,  and  compare  with  the  segment  of 
a  circle.  Do  we  speak  of  the  segment  of  a  circle  of  one  base?  Is  there 
any  reason  for  not  defining  the  segment  of  a  circle  so  that  we  can  speak 
of  a  segment  of  two  bases? 

8.  Is  the  distance  between  two  points  on  a  sphere  defined  so  that  it 
is  analogous  to  the  distance  between  two  points  in  a  plane? 


GENERAL  EXERCISES  425 

9.  Explain  why  it  is  well  to  treat  spherical  polygons  in  connection 
with  polyedral  angles.  State  the  relations  between  the  parts  of  a  spherical 
triangle  and  the  parts  of  the  corresponding  polyedral  angle. 

10.  State  several  theorems  about  spherical  triangles  that  are  analogous 
to  theorems  about  plane  triangles.  Is  a  plane  triangle  determined  when 
its  angles  are  known?  Is  a  spherical  triangle  determined  when  its  angles 
are  known? 

11.  What  is  a  polar  triangle?  What  are  some  relations  between  a 
spherical  triangle  and  its  polar  triangle?  State  a  striking  difference 
between  the  angles  of  a  spherical  triangle  and  those  of  a  plane  triangle. 

12.  How  many  points  determine  a  circle?     A  sphere? 

13.  How  many  points  on  a  sphere  determine  a  small  circle?  A  great 
circle? 

14.  Define  axis,  pole,  polar  distance. 

15.  If  a  great  circle  is  perpendicular  to  a  small  circle,  what  must  it 
contain?     Is  the  converse  of  this  true? 

16.  Give  an  outline  of  the  steps  in  determining  the  formulas  for  the 
area  of  a  sphere.  State  these  formulas,  and  state  when  each  is  the  best 
one  to  use. 

17.  Do  the  same  for  the  volume  of  a  sphere. 

18.  What  is  the  formula  for  the  area  of  a  zone?  A  lune?  A  spherical 
triangle?  Can  you  give  a  practical  application  of  the  use  of  each  of  these 
areas? 

19.  What  is  the  formula  for  the  volume  of  a  spherical  segment?  A 
wedge?  A  spherical  pyramid?  A  spherical  sector?  Can  you  give  a 
practical  application  of  the  use  of  each  of  these  volumes? 

20.  What  remarkable  relation  is  there  between  tlie  volumes  of  a 
cylinder,  sphere,  and  cone?     Is  there  a  similar  relation  between  their  areas? 

GENERAL  EXERCISES 

1.  The  volume  of  any  polyedron  circumscribed  about  a  sphere  is 
equal  to  the  area  of  its  surface  times  one-third  the  radius  of  the  sphere. 

2.  The  volumes  of  polyedrons  circumscribed  about  the  same  sphere 
are  to  each  other  as  their  surfaces. 

3.  What  is  the  locus  of  all  points  in  space  equally  distant  from  two 
given  points  and  at  a  distance  r  from  a  third  given  point? 

4.  Show  how  to  find  the  pole  of  a  given  circle  on  a  sphere. 

5.  Construct  a  circle  through  three  given  points  on  a  sphere. 


426  SOLID  GEOMETRY 

6.  Four  balls  of  equal  radii  r  are  placed  on  a  plane  and  touching  each 
other  in  such  a  way  that  their  centers  form  a  square.  A  fifth  ball  of  the 
same  radius  is  placed  upon  the  four  first  balls.  Find  the  distance  of  its 
center  from  the  plane. 

7.  A  cylinder  is  inscribed  in  a  sphere  of  radius  r  and  has  a  lateral 
area  equal  to  one-half  the  area  of  the  sphere.  Find  the  radius  of  the 
cylinder.  Ans.    %r\/~2. 

8.  A  zone  of  one  base  is  a  mean  proportional  between  the  remaining 
part  of  the  area  of  the  sphere  and  the  total  area  of  the  sphere.  How 
far  is  its  base  from  the  center  of  the  sphere? 

9.  A  spherical  segment  of  one  base  is  half  as  large  as  the  spherical  sector 
to  which  it  belongs.  If  the  radius  of  the  sphere  is  4  in.,  find  the  altitude 
of  the  segment. 

10.  The  inside  of  a  glass  is  in  the  form  of  a  right  circular  cone  whose 
vertex  angle  is  60°  and  the  diameter  of  whose  base  is  4  in.  The  glass  is 
filled  with  water  and  the  largest  sphere  that  can  be  immersed  in  the  water 
is  placed  in  the  glass.     Find  the  volume  of  water  remaining  in  the  glass. 

11.  An  irregular  portion,  but  less  than  half,  of  a  material  sphere,  is 
given.     Show  how  to  construct  the  diameter  of  the  sphere. 

12.  In  the  figure  of  §  813,  the  plane  MNP  is  tangent  to  the  sphere 
inscribed  in  the  cube.     If  the  edge  of  the  cube  is  a  find  DN. 

13.  Two  spheres  whose  radii  are  respectively  6 
in.  and  8  in.  have  their  centers  10  in.  apart.  Find 
the  volume  of  the  portion  common  to  the  two 
spheres.     This  is  the  form  of  a  spherical  lens. 

14.  A  hole  of  diameter  4  in.  was  bored  through 
the  center  of  a  sphere  of  diameter  12  in.  Find  the 
volume  of  the  part  cut  away. 

15.  A  sphere  is  inscribed  in  a  right  circular  cylinder  having  an  altitude 
equal  to  the  diameter  of  the  sphere.  Two  planes  parallel  to  the  base  of 
the  cylinder  cut  the  cylinder  and  the  sphere.  Prove  that  the  surface  of 
the  cylinder  lying  between  the  planes  equals  the  area  of  the  zone  between 
the  planes. 

16.  A  circle  is  circumscribed  about  an  equilateral  triangle.  Find  the 
ratios  of  the  areas  and  the  volumes  of  the  solids  formed  by  revolving  the 
triangle  and  the  circle  about  an  altitude  of  the  triangle  as  axis. 

17.  A  hollow  copper  sphere  used  as  a  float  in  water  weighs  10  oz.,  and 
has  a  diameter  of  5  in.     How  heavy  a  weight  will  it  support? 

Arts.     Less  than  27.9  oz. 


FORMULAS  FOR  REFERENCE  427 

FORMULAS   FOR   REFERENCE 

Lateral  area  of  prism.     S  =  pe,  S  =  ph. 

Lateral  area  of  circular  cylinder.     S  —  pe>  S  =  ch. 

Volume  of  rectangular  parallelepiped.     V  =  abh. 

Volume  of  parallelepiped.     V  =  Bh. 

Volume  of  prism  or  cylinder.     V  =  Bh. 

Volume  of  hollow  circular  cylinder.     V  =  \irh(D+d)(D  —  d). 

Lateral  area  of  regular  pyramid.     S  =  ^ps. 

Lateral  area  of  right  circular  cone.     S  =  ^cs  =  wrs. 

Total  area  of  right  circular  cone.     T  =  Trs+wr2  =  irr(s+r). 

Lateral  area  of  frustum  of  regular  pyramid.     S  =  ^s(P+p). 

Lateral  area  of  frustum  of  right  circular  cone.     S  =  ir(R+r)s. 

Volume  of  pyramid  or  cone.     V  =  ^Bh. 

Volume  of  circular  cone.     V  —  \irr2h. 

Volume  of  frustum  of  pyramid  or  cone. 

Vm%h(Bi+Bt+\fihBl). 

Volume  of  frustum  of  circular  cone. 

7  =  |7r/i(r12+r22+r1r2)  =-&7rh(d1*+d22+dld2). 

Volume  of  prismatoid.     7  =  fft(J?i+JB2+4Af). 

Area  of  a  sphere.     >S  =  47rr2. 

Area  of  a  zone.     Z  =  2irrh. 

u 
Area  of  a  lune.     L  —  — (47rr2). 
360 

Volume  of  a  sphere.     V  =  jfirr'6  =  f  7rd3  =  Jr& 

Volume  of  spherical  segment.     V  =  ^irh(ri2+r22)+^wh3. 

11 
Volume  of  spherical  wedge.     V = Of  Trr3) . 

Area  of  spherical  polygon.     A  = . 


428  SOLID  GEOMETRY 

USEFUL   NUMBERS 

1  cu.  ft.  of  water  weighs  62.5  lb.  (approx.)  =  1000  oz. 

1  gal.  of  water  weighs  8 J  lb.  (approx.). 

1  atmosphere  pressure  =  14.7  lb.  per  sq.  in.  =  2116.  lb.  per  sq.  ft 

1  atmosphere  pressure  =  760  mm.  of  mercury. 

A  column  of  water  2.3  ft.  high  =  a  pressure  of  1  lb.  per  sq.  in. 

1  Kg.  =  2.2  lb.  (approx.). 

1  gal.  =  231  cu.  in.  (by  law  of  Congress). 

1  cu.  ft.  =  7^  gal.  (approx.)  or  better  7.48  gal. 

1  cu.  ft.  =  4  bu.  (approx.). 

1  bbl.  =4.211-  cu.  ft.  (approx.). 

1  bu.  =  2150.42  cu.  in.  (by  law  of  Congress)  =  1.24446-  cu.  ft. 

1  bu.=f  cu.  ft.  (approx.). 

1  perch  =  24f  cu.  ft.  but  usually  taken  25  cu.  ft. 

1  in.  =  25.4  mm.  (approx.). 

1  m.  =  39.37  in.  (by  law  of  Congress). 

1  lb.  (avoirdupois)  =  7000  grains  (by  law  of  Congress). 

1  lb.  (troy  or  apothecaries)  =  5760  grains. 

7r  =  3.14159265358979+=3.1416  =  fff  =  3+  (all  approx.). 

V2=  1.4142136.  \/3  =  1.7320508. 

V5  =  2.2360680.  \/6  =  2.4494897. 

V^  1.2599210.  V^  =  1.4422496. 


SYLLABUS   OF   PLANE   GEOMETRY 

(FOR   REFERENCE   OR   REVIEW) 

Chapter   I.     Straight-Line  Figures 

PRELIMINARY   PROPOSITIONS   AND   AXIOMS 

§  19.  Experiment.  If  two  triangles  have  two  angles  and  the  included 
side  of  one  equal  respectively  to  two  angles  and  the  included  side  of  the 
other,  the  triangles  are  equal. 

§  21.  In  equal  triangles,  the  corresponding  sides  are  equal  and  the 
corresponding  angles  are  equal. 

§  23.  Experiment.  If  two  triangles  have  two  sides  and  the  included 
angle  of  one  equal  respectively  to  two  sides  and  the  included  angle  of  the 
other,  the  triangles  are  equal. 

§  25.  Construction.  To  construct  a  triangle  when  the  three  sides 
are  given. 

§  27.  Experiment.  If  two  triangles  have  the  three  sides  of  one  equal 
respectively  to  the  three  sides  of  the  other,  the  triangles  are  equaL 

§  30.  Experiment.  The  sum  of  the  angles  of  a  triangle  is  equal  to 
180°. 

§  32.  If  two  triangles  have  two  angles  of  one  equal  respectively  to 
two  angles  of  the  other,  the  third  angles  are  equal. 

§  35.  Experiment.  If  one  straight  line  intersects  another  straight 
line,  the  vertical  angles  are  equal. 

§  49.  If  two  adjacent  angles  have  their  exterior  sides  in  the  same 
straight  line,  they  are  supplementary. 

§  51.  If  two  adjacent  angles  are  supplementary,  their  exterior  sides 
are  in  the  same  straight  line. 

§  53.     Complements  of  the  same  angle  or  of  equal  angles  are  equal. 

§  55.     Supplements  of  the  same  angle  or  of  equal  angles  are  equal. 

§  58.  If  one  of  the  angles  formed  by  two  intersecting  lines  is  a  right 
angle,  the  other  three  angles  also  are  right  angles. 

§  59.  The  sum  of  all  the  angles  about  a  point  is  equal  to  360°,  or  four 
right  angles. 

§  60.  The  sum  of  all  the  angles  about  a  point,  on  the  same  side  of  a 
straight  line  passing  through  the  point,  is  equal  to  180°. 

§  61.     All  right  angles  are  equal. 

§  62.     All  straight  angles  are  equal. 

429 


430  SOLID   GEOMETRY 

§  63.     A  right  angle  is  half  a  straight  angle. 

§  65.  One  straight  line,  and  only  one,  can  be  drawn  from  one  point 
to  another. 

§  66.     Two  points  determine  a  straight  line. 

§  67.     A  straight  line  may  be  produced  indefinitely  in  both  directions. 

§  68.  Any  number  of  straight  lines  can  be  drawn  through  a  given 
point. 

§  70.     Two  straight  lines  can  intersect  in  only  one  point. 

§  71.     Two  intersecting  lines  determine  a  point. 

§  74.  At  a  given  point  in  a  given  straight  line,  only  one  perpendicular 
can  be  drawn  to  that  line. 

§  76.  Only  one  perpendicular  can  be  drawn  to  a  given  straight  line 
from  a  given  external  point . 

§  77.  Through  a  given  point,  one  straight  line,  and  only  one,  can  be 
drawn  perpendicular  to  a  given  straight  line. 


TRIANGLES 

§  84.  Problem.  To  construct  an  equilateral  triangle  when  a  side  is 
given. 

§  85.  Problem.  To  construct  an  isosceles  triangle  when  the  base 
and  one  of  the  equal  sides  are  given. 

§  86.  Theorem.  In  an  isosceles  triangle,  the  angles  opposite  the 
equal  sides  are  equal. 

§  87.     Theorem.     An  equilateral  triangle  is  also  equiangular. 

§  88.     Theorem.     Each  angle  of  an  equilateral  triangle  is  equal  to  60°. 

§  89.  Theorem.  If  two  angles  of  a  triangle  are  equal  the  sides  oppo- 
site the  equal  angles  are  equal  and  the  triangle  is  isosceles. 

§  90.     Theorem.     An  equiangular  triangle  is  also  equilateral. 

§  91.  Theorem.  The  sum  of  the  acute  angles  of  a  right  triangle  is 
equal  to  90°,  or  one  right  angle. 

§  92.  Theorem.  If  two  right  triangles  have  one  acute  angle  of  one 
equal  to  one  acute  angle  of  the  other,  the  other  acute  angles,  also,  are  equal. 

§  93.  Theorem.  Each  acute  angle  of  an  isosceles  right  triangle  is 
equal  to  45°. 

§  97.  If  two  right  triangles  have  a  side  and  an  acute  angle  of  one 
equal  respectively  to  the  corresponding  side  and  the  corresponding  acute 
angle  of  the  other,  the  triangles  are  equal. 

§  98.  Theorem.  If  two  triangles  have  a  side  and  any  two  angles  of 
one  equal  respectively  to  the  corresponding  side  and  the  two  correspond- 
ing angles  of  the  other,  the  triangles  are  equal. 

§  101.  If  two  right  triangles  have  two  sides  of  one  equal  respectively 
to  two  corresponding  sides  of  the  other,  the  triangles  are  equal. 


SYLLABUS  OF  PLANE  GEOMETRY  431 

AXIOMS   OF  EQUALITY 

§  104.  Axiom.  Quantities  which  are  equal  to  the  same  quantity  or 
to  equal  quantities  are  equal  to  each  other. 

§  105.     Axiom.     If  equals  are  added  to  equals,  the  sums  are  equal. 

§  106.  Axiom.  If  equals  are  subtracted  from  equals,  the  remainders 
are  equal. 

§  107.  Axiom.  If  equals  are  multiplied  by  equals,  the  products  are 
equal. 

§  108.  Axiom.  If  equals  are  divided  by  equals,  the  quotients  are 
equal.     The  divisor  must  not  be  zero. 

§  109.     Axiom.     The  whole  is  equal  to  the  sum  of  all  its  parts. 

§  110.     Axiom.     The  whole  is  greater  than  any  of  its  parts. 

§  111.  Axiom.  A  quantity  may  be  substituted  for  its  equal  in  any 
operation. 

§  113.  Theorem.  If  two  triangles  have  the  three  sides  of  one  equal 
respectively  to  the  three  sides  of  the  other,  the  triangles  are  equal. 

PARALLEL  LINES 

§  115.  Theorem.  Straight  lines  in  the  same  plane  which  are  per- 
pendicular to  the  same  straight  line  can  never  meet. 

§  117.  Theorem.  Straight  lines  in  the  same  plane  which  are  per- 
pendicular to  the  same  straight  line  are  parallel. 

§  118.  Axiom.  Through  a  given  point  only  one  straight  line  can  be 
drawn  parallel  to  another  straight  line. 

§  119.  Axiom.  Straight  lines  in  the  same  plane  which  are  not  par- 
allel will  meet  if  sufficiently  prolonged. 

§  122.     Experiment.     If  two  parallel  lines  are  cut  by  a  transversal: 

(1)  The  alternate  interior  angles  are  equal. 

(2)  The  alternate  exterior  angles  are  equal. 

(3)  The  exterior  interior  angles  are  equal. 

(4)  The  consecutive  interior  angles  are  supplementary. 

(5)  The  consecutive  exterior  angles  are  supplementary. 

§  123.  Theorem.  If  a  straight  line  is  perpendicular  to  one  of  two 
parallel  lines,  it  is  perpendicular  to  the  other  also. 

§  124.  Theorem.  Two  straight  lines  parallel  to  a  third  straight 
line  are  parallel  to  each  other. 

§  127.  Theorem.  If  two  straight  lines  in  the  same  plane  are  cut  by 
a  transversal,  making  the  alternate  interior  angles  equal,  the  lines  are 
parallel. 

§  128.  Theorem.  If  two  straight  lines  in  the  same  plane  are  cut  by  a 
transversal,  making  the  exterior  interior  angles  equal,  the  lines  are  parallel. 


432  SOLID   GEOMETRY 

§  129.  Theorem.  If  two  straight  lines  in  the  same  plane  are  cut  by 
a  transversal,  making  the  alternate  exterior  angles  equal,  the  lines  are 
parallel. 

§  130.  Theorem.  If  two  straight  lines  in  the  same  plane  are  cut  by 
a  transversal,  making  the  consecutive  exterior  angles  supplementary,  the 
lines  are  parallel. 

§  131.  Theorem.  If  two  straight  lines  in  the  same  plane  are  cut  by 
a  transversal,  making  the  consecutive  interior  angles  supplementary,  the 
lines  are  parallel. 

§  132.  Theorem.  If  two  parallel  lines  are  cut  by  a  transversal,  the 
alternate  interior  angles  are  equal. 

§  133.  Theorem.  If  two  parallel  lines  are  cut  by  a  transversal,  the 
exterior  interior  angles  are  equal. 

§  134.  Theorem.  If  two  parallel  lines  are  cut  by  a  transversal,  the 
alternate  exterior  angles  are  equal. 

§  135.  Theorem.  If  two  parallel  lines  are  cut  by  a  transversal,  the 
consecutive  interior  angles  are  supplementary. 

§  136.  Theorem.  If  two  parallel  lines  are  cut  by  a  transversal,  the 
consecutive  exterior  angles  are  supplementary. 

§  138.  Theorem.  If  two  angles  have  their  sides  parallel,  right  side 
to  right  side,  and  left  side  to  left  side,  the  angles  are  equal. 

§  139.  Theorem.  If  two  angles  have  their  sides  parallel,  right  side 
to  left  side,  and  left  side  to  right  side,  the  angles  are  supplementary. 

§  140.  Theorem.  If  two  angles  have  their  sides  perpendicular,  right 
side  to  right  side,  and  left  side  to  left  side,  the  angles  are  equal. 

§  141.  Theorem.  If  two  angles  have  their  sides  perpendicular,  right 
side  to  left  side  and  left  side  to  right  side,  the  angles  are  supplementary. 

§  143.  Theorem.  The  sum  of  the  angles  of  a  triangle  is  equal  to 
180°,  or  two  right  angles. 

§  144.  Theorem.  An  exterior  angle  of  a  triangle  is  equal  to  the  sum 
of  the  opposite  interior  angles,  and  is  therefore  greater  than  either  of  them. 

§  145.  Theorem.  In  a  triangle,  there  can  be  only  one  right  angle,  or 
one  obtuse  angle. 

QUADRILATERALS 

§  154.  Theorem.  A  diagonal  divides  a  parallelogram  into  two  equal 
triangles. 

§  155.  Theorem.  In  any  parallelogram,  the  opposite  sides  are  equal, 
and  the  opposite  angles  are  equal. 

§  156.     Theorem.     The  diagonals  of  a  parallelogram  bisect  each  other. 

§  157.  Theorem.  Any  two  consecutive  angles  of  a  parallelogram 
are  supplementary. 


SYLLABUS  OF  PLANE  GEOMETRY  433 

§  168.     Theorem.     Parallels  cut  off  by  two  parallel  lines  are  equal. 

§  169.  Theorem.  If  the  opposite  sides  of  a  quadrilateral  are  equal, 
the  figure  is  a  parallelogram. 

§  160.  Theorem.  If  two  opposite  sides  of  a  quadrilateral  are  equal 
and  parallel,  the  figure  is  a  parallelogram. 

§  161.  Theorem.  If  the  diagonals  of  a  quadrilateral  bisect  each 
other,  the  figure  is  a  parallelogram. 

§  162.  Theorem.  The  line  joining  the  middle  points  of  two  opposite 
sides  of  a  parallelogram  is  parallel  to  and  equal  to  each  of  the  other  two 
sides. 

§  163.     Theorem.     The  diagonals  of  a  rectangle  are  equal. 

§  164.  Theorem.  If  a  parallelogram  has  equal  diagonals,  it  is  a 
rectangle. 

§  165.  Theorem.  The  diagonals  of  a  square  are  perpendicular  to 
each  other,  and  bisect  the  angles  of  the  square. 

POLYGONS 

§  170.     Theorem.     The  sum  of  the  interior  angles  of  any  polygon  is 

equal  to  180°  multiplied  by  two  less  than  the  number  of  sides. 

§  171.     Theorem.     In  an  equiangular  polygon,  each  angle  is  equal  to 

(tt-2)180°        (n-2)2rt.^ 

,  or . 

n  n 

§  172.     Theorem.     The  sum  of  the  exterior  angles  of  a  polygon,  formed 

by  prolonging  one  side  at  each  vertex,  is  360°. 

INEQUALITIES 

§  173.  Axiom.  If  equals  are  added  to  unequals,  the  sums  are  unequal 
in  the  same  order. 

§  174.  Axiom.  If  equals  are  subtracted  from  unequals,  the  remainders 
are  unequal  in  the  same  order. 

§  175.  Axiom.  If  unequals  are  multiplied  by  positive  equals,  the 
products  are  unequal  in  the  same  order. 

§  176.  Axiom.  If  unequals  are  divided  by  positive  equals,  the  quo- 
tients are  unequal  in  the  same  order. 

§  177.  Axiom.  If  unequals  are  added  to  unequals  in  the  same  order, 
the  sums  are  unequal  in  that  order. 

§  178.  Axiom.  If  unequals  are  subtracted  from  equals,  the  remain- 
ders are  unequal  in  the  reverse  order. 

§  179.  Axiom.  If  the  first  of  three  quantities  is  greater  than  the 
second  and  the  second  is  greater  than  the  third,  then  the  first  is  greater 
than  the  third, 


434  SOLID  GEOMETRY 

§  181.  Axiom.  A  straight  line  is  the  shortest  line  that  can  be  drawn 
from  one  point  to  another. 

§  182.  Theorem.  The  sum  of  two  sides  of  a  triangle  is  greater  than 
the  third  side.  The  difference  between  two  sides  of  a  triangle  is  less  than 
the  third  side. 

§  183.  Theorem.  If  two  sides  of  a  triangle  are  unequal,  the  angles 
opposite  these  sides  are  unequal,  and  the  greater  angle  is  opposite  the 
greater  side. 

§  184.  Theorem.  If  two  angles  of  a  triangle  are  unequal,  the  sides 
opposite  these  angles  are  unequal,  and  the  greater  side  is  opposite  the 
greater  angle. 

§  185.  Theorem.  A  perpendicular  is  the  shortest  line  that  can  be 
drawn  from  a  point  to  a  straight  line. 

§  186.  Theorem.  If  two  straight  lines  are  drawn  from  a  point  in  a 
perpendicular  to  a  given  line,  cutting  off  on  the  line  unequal  lengths  from 
the  foot  of  the  perpendicular,  the  more  remote  is  the  greater. 

DISTANCE 

§  191.     Theorem.     Two  parallel  lines  are  everywhere  equally  distant. 

§  192.  Theorem.  If  a  line  bisects  one  side  of  a  triangle  and  is  par- 
allel to  a  second  side,  it  bisects  the  third  side. 

§  193.  Theorem.  The  middle  point  of  the  hypotenuse  of  a  right 
triangle  is  equidistant  from  the  three  vertices. 

§  194.  Theorem.  If  one  acute  angle  of  a  right  triangle  is  twice  the 
other,  the  hypotenuse  is  twice  the  shorter  side. 

§.  196.  Theorem.  The  line  joining  the  middle  points  of  two  sides  of 
a  triangle  is  parallel  to  the  third  side  and  is  equal  to  one-half  the  third  side. 

§  197.  Theorem.  If  three  or  more  parallel  lines  intercept  equal 
parts  on  one  transversal,  they  intercept  equal  parts  on  every  transversal. 

§  198.  Theorem.  The  median  of  a  trapezoid  is  parallel,  to  the  bases 
and  is  equal  to  half  the  sum  of  the  bases. 

§  199.  Theorem.  All  points  in  the  bisector  of  an  angle  are  equi- 
distant from  the  sides  of  the  angle. 

§  200.  Theorem.  All  points  equidistant  from  the  sides  of  aji  angle 
are  in  the  bisector  of  the  angle. 

§  205.  Theorem.  All  points  in  the  perpendicular  bisector  of  a  straight 
line  are  equidistant  from  the  extremities  of  the  line. 

§  206.  Theorem.  All  points  equidistant  from  the  extremities  of  a 
straight  line  lie  in  the  perpendicular  bisector  of  the  line. 

§  207.  Theorem.  Two  points,  each  equidistant  from  the  extremities 
of  a  straight  line,  determine  the  perpendicular  bisector  of  the  line. 


SYLLABUS  OF  PLANE  GEOMETRY  435 

OTHER  THEOREMS 

§  209.  Theorem.  The  bisector  of  the  angles  of  a  triangle  pass  through 
a  common  point,  which  is  equidistant  from  the  sides  of  the  triangle. 

§  210.  Theorem.  The  perpendicular  bisectors  of  the  three  sides  of  a 
triangle  pass  through  a  common  point,  which  is  equidistant  from  the  three 
vertices  of  the  triangle. 

§  211.  Theorem.  The  medians  of  a  triangle  pass  through  a  common 
point,  which  is  two-thirds  of  the  length  of  each  median  from  its  vertex. 

§  213.  Theorem.  The  three  altitudes  of  a  triangle  pass  through  a 
common  point. 

§  247.  Theorem.  Two  triangles  are  congruent  if  two  sides  and  the 
included  angle  of  one  are  equal  respectively  to  two  sides  and  the  included 
angle  of  the  other. 

§  248.  Theorem.  Two  triangles  are  congruent  if  two  angles  and  the 
included  side  of  one  are  equal  respectively  to  two  angles  and  the  included 
side  of  the  other. 

§  249.  Theorem.  Two  parallelograms  are  congruent  if  they  have 
two  adjacent  sides  and  the  included  angle  of  one  equal  respectively  to 
two  adjacent  sides  and  the  included  angle  of  the  other. 

§  251.  Theorem.  If  two  angles  of  a  triangle  are  unequal,  the  sides 
opposite  these  angles  are  unequal,  the  greater  side  being  opposite  the 
greater  angle. 

§  252.  Theorem.  Only  one  perpendicular  can  be  drawn  to  a  given 
line  at  a  given  point  in  the  line. 

§  253.  Theorem.  Only  one  perpendicular  can  be  drawn  to  a  given 
line  from  a  given  external  point. 

§  254.  Theorem.  If  two  parallel  lines  are  cut  by  a  transversal,  the 
alternate  interior  angles  are  equal. 

§  255.  Theorem.  If  two  parallel  lines  are  cut  by  a  transversal,  the 
exterior  interior  angles  are  equal. 

§  256.  Theorem.  If  two  parallel  lines  are  cut  by  a  transversal,  the 
consecutive  interior  angles  are  supplementary. 

§  257.  Theorem.  If  two  parallel  lines  are  cut  by  a  transversal,  the 
alternate  exterior  angles  are  equal. 

§  258.  Theorem.  If  two  triangles  have  two  sides  of  one  equal 
respectively  to  two  sides  of  the  other  but  the  included  angle  of  the  first 
triangle  greater  than  the  included  angle  of  the  second,  then  the  third  side 
of  the  first  is  greater  than  the  third  side  of  the  second. 

§  259.  Theorem.  If  two  triangles  have  two  sides  of  one  equal  to 
two  sides  of  the  other  but  the  third  side  of  the  first  greater  than  the  third 
side  of  the  second,  then  the  angle  opposite  the  third  side  of  the  first  is 
greater  than  the  angle  opposite  the  third  side  of  the  second. 


436  SOLID  GEOMETRY 


Chapter  II.     The  Circle 

§  272.  Experiment.  Equal  central  angles  in  the  same  circle  or  in 
equal  circles  intercept  equal  arcs.  Conversely:  Equal  arcs  are  inter- 
cepted by  equal  central  angles. 

§  273.  Of  two  unequal  central  angles,  in  the  same  circle  or  in  equal 
circles,  the  greater  central  angle  intercepts  the  greater  arc. 

Conversely:  Of  two  unequal  arcs,  the  greater  arc  is  intercepted  by 
the  greater  central  angle. 

ARCS   AND    CHORDS 

§  275.  Theorem.  In  the  same  circle  or  in  equal  circles,  equal  chords 
subtend  equal  arcs.  Conversely:  Equal  arcs  are  subtended  by  equal 
chords. 

§  276.  Theorem.  A  diameter  of  a  circle  is  greater  than  any  other 
chord. 

§  277.  Theorem.  A  diameter  perpendicular  to  a  chord  bisects  the 
chord  and  the  arc  subtended  by  it. 

§  278.     Theorem.     A  diameter  bisects  the  circle. 

§  279.  Theorem.  A  diameter  which  bisects  a  chord  is  perpendicu- 
lar to  the  chord. 

§  280.  Theorem.  The  perpendicular  bisector  of  a  chord  passes 
through  the  center  of  the  circle. 

§  281.  Theorem.  In  the  same  circle  or  in  equal  circles,  equal  chords 
are  equidistant  from  the  center.  Conversely:  Chords  equidistant  from 
the  center  are  equal. 

§  282.  The  distance  from  the  center  of  a  circle  to  any  point  without 
the  circle  is  greater  than  the  radius,  and  the  distance  from  the  center  to 
any  point  within  the  circle  is  less  than  the  radius. 

§  283.  A  point  is  without  a  circle  when  its  distance  from  the  center 
is  greater  than  the  radius.  Also,  a  point  is  within  a  circle  when  its  dis- 
tance from  the  center  is  less  than  the  radius. 

TANGENTS   AND   SECANTS 

§  284.  Theorem.  A  perpendicular  to  a  radius  at  its  outer  extremity 
can  touch  the  circle  at  only  one  point. 

§  287.  Theorem.  A  straight  line  perpendicular  to  a  radius  at  the 
point  where  the  radius  meets  the  circle  is  tangent  to  the  circle. 

§  288.  Theorem.  The  radius  drawn  to  the  point  of  tangency  is  per- 
pendicular to  the  tangent. 

§  289.     Theorem.     Parallel  lines  intercept  equal  arcs  on  a  circle. 


SYLLABUS  OF  PLANE  GEOMETRY  437 

J  290.  Theorem.  Tangents  to  a  circle  from  an  external  point  are 
equal,  and  make  equal  angles  with  the  line  joining  that  point  with  the 
center. 

§  295.  Theorem.  The  line  of  centers  of  two  intersecting  circles  is 
the  perpendicular  bisector  of  their  common  chord. 

§  296.  Theorem.  If  two  circles  are  tangent  to  each  other,  the  line 
of  centers  passes  through  the  point  of  contact. 

ANGLE   MEASUREMENT 

§  303.  Theorem.  The  arcs  formed  by  two  perpendicular  diameters 
are  equal. 

§  305.  Experiment.  A  central  angle  equals  in  degrees  its  intercepted 
arc. 

§  306.  Theorem.  An  inscribed  angle  equals  in  degrees  one-half  its 
intercepted  arc. 

§  307.     Theorem.     Angles  inscribed  in  the  same  arc  are  equal. 

§  308.     Theorem.     An  angle  inscribed  in  a  semicircle  is  a  right  angle. 

§  309.  Theorem.  An  angle  formed  by  two  intersecting  chords  equals 
in  degrees  one-half  the  sum  of  the  intercepted  arcs. 

§  310.  Theorem.  An  angle  formed  by  a  tangent  and  a  chord  equals 
in  degrees  one-half  the  intercepted  arc. 

§  311.  Theorem.  An  angle  formed  by  two  secants,  two  tangents,  or 
a  tangent  and  a  secant,  equals  in  degrees  one-half  the  difference  of  the 
intercepted  arcs. 

PROBLEMS   OF   CONSTRUCTION 

§  312.  Problem.  To  construct  a  perpendicular  to  a  given  straight 
line  from  a  given  external  point. 

§  313.  Problem.  To  construct  a  perpendicular  to  a  given  straight 
line  at  a  given  point  in  the  line. 

§  314.     Problem.     To  bisect  a  given  straight  line. 

§  315.     Problem.     To  bisect  a  given  arc. 

§  316.     Problem.     To  circumscribe  a  circle  about  a  given  triangle. 

§  317.  Problem.  To  construct  a  circle  through  three  points  which 
are  not  in  the  same  straight  line. 

§  318.  Theorem.  Three  points  not  in  the  same  straight  line  deter- 
mine a  circle. 

§  319.     Problem.     To  bisect  a  given  angle. 

§  320.  Problem.  From  a  given  point  in  a  given  line,  to  draw  a  line 
making  an  angle  equal  to  a  given  angle. 

§  321.     Problem.     Lines  may  be   added,  subtracted,  and   multiplied. 

§  322.     Problem'.     A  line  may  be  divided  into  equal  parts. 


438  SOLID  GEOMETRY 

§  323.  Problem.  To  construct  a  triangle  when  two  angles  and  the 
included  side  are  given. 

§  324.  Problem.  To  construct  a  triangle  when  two  sides  and  the 
included  angle  are  given. 

§  325.     Problem.     To  inscribe  a  circle  in  a  given  triangle. 

§  326.     Problem.     To  construct  a  tangent  to  a  given  circle. 

§  327.  Problem.  With  a  given  straight  line  as  a  chord,  to  construct 
an  arc  of  a  circle  in  which  a  given  angle  may  be  inscribed. 

§  329.  Problem.  To  construct  a  triangle  when  two  sides  and  the 
angle  opposite  one  of  them  are  given. 


Chapter  III.     Area  of  Polygons 

MEASURING 

§  333.     Axiom.     Every  geometric  magnitude  has  a  numerical  measure. 

§  346.  The  area  of  a  rectangle  is  equal  to  the  product  of  its  base  and 
altitude. 

§  347.     Theorem.     The  area  of  a  square  equals  the  square  of  its  side. 

§  348.  Theorem.  The  areas  of  two  rectangles  are  to  each  other  as 
the  products  of  their  bases  and  altitudes. 

§  349.  Theorem.  The  areas  of  two  rectangles  having  equal  bases 
are  to  each  other  as  their  altitudes. 

§  350.  Theorem.  The  areas  of  two  rectangles  having  equal  altitudes 
are  to  each  other  as  their  bases. 

§  351.  Theorem.  Two  rectangles  having  equal  bases  and  equal 
altitudes  are  equal  in  area. 

§  354.  Theorem.  Any  parallelogram  is  equivalent  to  a  rectangle 
that  has  its  base  and  altitude  equal  respectively  to  the  base  and  altitude 
of  the  parallelogram. 

§  355.  Theorem.  The  area  of  a  parallelogram  is  equal  to  the  prod- 
uct of  its  base  and  altitude. 

§  356.  Theorem.  The  altitude  of  a  parallelogram  is  equal  to  its 
area  divided  by  its  base. 

§  357.  Theorem.  The  base  of  a  parallelogram  is  equal  to  its  area 
divided  by  its  altitude. 

§  358.  Theorem.  Parallelograms  having  equal  bases  and  equal  alti- 
tudes are  equivalent. 

§  359.  Theorem.  Two  parallelograms  having  equal  bases  are  to 
each  other  as  their  altitudes. 

§  360.  Theorem.  Two  parallelograms  having  equal  altitudes  are  to 
each  other  as  their  bases. 


SYLLABUS  OF   PLANE   GEOMETRY  439 

§  361.  Theorem.  Any  two  parallelograms  are  to  each  other  as  the 
products  of  their  bases  and  their  altitudes. 

§  362.  Theorem.  The  area  of  a  triangle  is  equal  to  half  the  product 
of  its  base  and  its  altitude. 

§  363.  Theorem.  The  area  of  a  triangle  equals  half  the*area  of  a 
parallelogram  having  the  same  base  and  altitude  as  the  triangle. 

§  364.  Theorem.  The  base  of  a  triangle  equals  twice  its  area  divided 
by  its  altitude. 

§  365.  Theorem.  The  altitude  of  a  triangle  equals  twice  its  area 
divided  by  its  base. 

§  366.  Theorem.  Triangles  having  equal  bases  and  equal  altitudes 
are  equivalent. 

§  367.  Theorem.  Two  triangles  having  equal  bases  are  to  each 
other  as  their  altitudes. 

§  368.  Theorem.  Two  triangles  having  equal  altitudes  are  to  each 
other  as  their  bases. 

§  369.  Theorem.  Any  two  triangles  are  to  each  other  as  the  prod- 
ucts of  their  bases  and  their  altitudes. 

§  370.  Theorem.  The  area  of  a  trapezoid  is  equal  to  half  the  prod- 
uct of  its  altitude  and  the  sum  of  its  bases. 

§  371.  Theorem.  The  area  of  a  trapezoid  equals  the  product  of  its 
altitude  and  its  median. 

§  375.  Theorem.  Two  triangles  that  have  an  angle  of  one  equal  to 
an  angle  of  the  other  are  to  each  other  as  the  products  of  the  sides  includ- 
ing the  equal  angles. 

§  376.  Theorem.  The  square  constructed  on  the  hypotenuse  of  a 
right  triangle  is  equivalent  to  the  sum  of  the  squares  constructed  on  the 
other  two  sides. 

§  377.  Theorem.  The  square  on  either  side  of  a  right  triangle  is 
equivalent  to  the  square  on  the  hypotenuse  minus  the  square  on  the  other 
side. 

PROJECTION 

§  380.  Theorem.  In  any  obtuse  triangle,  the  square  of  the  side 
opposite  the  obtuse  angle  is  equivalent  to  the  sum  of  the  squares  of  the 
other  two  sides,  plus  twice  the  product  of  one  of  those  sides  and  the  pro- 
jection of  the  other  side  upon  it. 

§  381.  Theorem.  In  any  triangle,  the  square  of  a  side  opposite  an 
acute  angle  is  equivalent  to  the  sum  of  the  squares  of  the  other  two  sides, 
minus  twice  the  product  of  one  of  these  sides  and  the  projection  of  the 
other  side  upon  it. 


440  SOLID   GEOMETRY 

§  382.  Theorem.  An  angle  of  a  triangle  is  acute,  right,  or  obtuse 
according  as  the  square  of  the  opposite  side  is  less  than,  equal  to,  or  greater 
than  the  sum  of  the  squares  of  the  other  two  sides. 

§  383.     Theorem.     If  any  median  of  a  triangle  be  drawn  to  a  side: 

(1)  The  sum  of  the  squares  of  the  other  two  sides  is  equal  to  twice  the 
square  of  the  median  plus  twice  the  square  of  half  the  bisected  side. 

(2)  The  difference  of  the  squares  of  the  other  two  sides  is  equal  to 
twice  the  product  of  the  bisected  side  and  the  projection  of  the  median 
upon  that  side. 

TRANSFORMATIONS   AND    CONSTRUCTIONS 

§  385.  Problem.  To  construct  a  square  that  is  equivalent  to  the 
sum  of  two  given  squares. 

§  386.  Problem.  To  construct  a  square  that  is  equivalent  to  the 
difference  of  two  given  squares. 

§  388.  Problem.  To  transform  a  given  quadrilateral  into  an  equiv- 
alent triangle. 

§  389.  Problem.  To  transform  a  given  polygon  into  an  equivalent 
triangle. 

§  390.  Problem.  To  construct  a  square  equivalent  to  a  given  rec- 
tangle or  parallelogram. 

§  391.     Problem.     To  construct  a  square  equivalent  to  a  given  triangle. 

§  392.     Problem.     To  construct  a  square  equivalent  to  a  given  polygon. 


Chapter  IV.     Proportion  and  Similarity 
THEOREMS    OF   PROPORTION 

§  398.  Theorem.  If  four  numbers  form  a  proportion,  the  product  of 
the  extremes  equals  the  product  of  the  means. 

§  399.  Theorem.  The  mean  proportional  between  two  numbers 
equals  the  square  root  of  their  product. 

§  400.  Theorem.  If  the  two  antecedents  of  a  proportion  are  equal, 
the  two  consequents  are  equal. 

§  401.  Theorem.  If  three  terms  of  one  proportion  are  equal  respec- 
tively to  three  corresponding  terms  of  another  proportion,  the  fourth 
terms  are  equal. 

§  402.  Theorem.  If  the  product  of  two  numbers  equals  the  product 
of  two  other  numbers,  either  two  may  be  made  the  means  and  the  other 
two  the  extremes  of  a  proportion. 

§  403.  Theorem.  In  a  series  of  equal  ratios,  the  sum  of  the  anteced- 
ents is  to  the  sum  of  the  consequents  as  any  antecedent  is  to  its  consequent. 


SYLLABUS   OF  PLANE  GEOMETRY  441 

§  404.  Theorem.  If  four  numbers  form  a  proportion,  they  are  in 
proportion  by  inversion;  that  is,  the  second  term  is  US  the  first  as  the 
fourth  is  to  the  third. 

§  405.  Theorem.  If  four  numbers  form  a  proportion,  they  are  in 
proportion  by  alternation;  that  is,  the  first  term  is  to  the  third  as  the 
second  is  to  the  fourth. 

§  406.  Theorem.  If  four  numbers  form  a  proportion,  they  are  in 
proportion  by  addition;  that  is,  the  sum  of  the  first  two  terms  is  to  the 
second  as  the  sum  of  the  last  two  terms  is  to  the  fourth. 

§  407.  Theorem.  If  four  numbers  form  a  proportion,  they  are  in 
proportion  by  subtraction;  that  is,  the  difference  of  the  first  two  terms  is 
to  the  second  as  the  difference  of  the  last  two  terms  is  to  the  fourth. 

§  408.  Theorem.  If  four  numbers  form  a  proportion,  they  arc  in 
proportion  by  addition  and  subtraction;  that  is,  the  sum  of  the  first  two 
is  to  their  difference  as  the  sum  of  the  last  two  is  to  their  difference. 

PROPORTIONAL   LINES 

§  411.  Theorem.  A  straight  line  parallel  to  one  side  of  a  triangle 
divides  the  other  two  sides  in  the  same  ratio. 

§  412.  Theorem.  If  two  straight  lines  are  cut  by  a  series  of  parallel 
straight  lines,  the  corresponding  segments  are  proportional. 

§  413.  Theorem.  A  line  which  divides  two  sides  of  a  triangle  in  the 
same  ratio  is  parallel  to  the  third  side. 

§  414.  Problem.  To  construct  the  fourth  proportional  to  three  given 
lines. 

§  415.  Problem.  To  divide  a  given  straight  line  into  parts  that  are 
in  a  given  ratio. 

§  416.  Problem.  .To  divide  a  given  straight  line  into  parts  propor- 
tional to  any  number  of  given  lines. 

§  417.  Theorem.  The  bisector  of  an  angle  of  a  triangle  divides  the 
opposite  side  into  two  parts  which  are  proportional  to  the  adjacent  sides. 

SIMILAR   TRIANGLES 

§  421.  Theorem.  Two  triangles  similar  to  the  same  triangle  are 
similar  to  each  other.  i 

§  422.  Theorem.  Two  triangles  are  similar  if  two  angles  of  one  are 
equal  respectively  to  two  angles  of  the  other. 

§  423.  Theorem.  Two  right  triangles  are  similar  if  an  acute  angle 
of  one  is  equal  to  an  acute  angle  of  the  other. 

§  424.  Theorem.  Two  isosceles  triangles  are  similar  if  the  angle  at 
the  vertex,  or  a  base  angle,  of  one  equals  the  corresponding  angle  of  the 
other. 


442  SOLID  GEOMETRY 

§  425.     Theorem.     Two  equilateral  triangles  are  similar. 

§  426.  Theorem.  If  the  corresponding  angles  of  two  triangles  have 
their  sides  respectively  parallel,  right  side  to  right  side  and  left  side  to  left 
side,  the  triangles  are  similar. 

§  427.  Theorem.  If  the  corresponding  angles  of  two  triangles  have 
their  sides  respectively  perpendicular,  right  side  to  right  side  and  left  side 
to  left  side,  the  triangles  are  similar. 

§  428.  Theorem.  The  corresponding  sides  of  similar  triangles  are 
proportional. 

§  430.  Theorem.  If  two  triangles  have  their  corresponding  sides 
proportional,  they  are  similar. 

§  431.  Theorem.  If  two  triangles  have  an  angle  of  one  equal  to  an 
angle  of  the  other  and  the  including  sides  proportional,  they  are 
similar.  . 

§  432.  Theorem.  The  areas  of  two  similar  triangles  are  in  the  same 
ratio  as  the  squares  of  any  two  corresponding  sides. 

§  434.  .  Theorem.  In  any  right  triangle,  if  the  altitude  is  drawn  from 
the  vertex  of  the  right  angle  to  the  hypotenuse: 

(1)  The  two  triangles  thus  formed  are  similar  to  the  given  triangle  and 
to  each  other. 

(2)  The  altitude  is  the  mean  proportional  between  the  segments  of  the 
hypotenuse. 

§  435.  Theorem.  Each  side  about  the  right  angle  is  the  mean  pro- 
portional between  the  hypotenuse  and  its  adjacent  segment. 

§  436.  Theorem.  The  two  segments  of  the  hypotenuse  and  the 
squares  of  the  two  sides  are  proportionals. 

§  437.  Theorem.  The  square  of  the  hypotenuse  is  to  the  square  of 
either  side  as  the  hypotenuse  is  to  the  segment  of  the  hypotenuse  adjacent 
to  that  side. 

§  438.  Theorem.  The  square  of  the  hypotenuse  equals  the  sum  of 
the  squares  of  the  two  sides. 

§  439.  Theorem.  A  perpendicular  drawn  to  a  diameter  from  any 
point  on  the  circle  is  a  mean  proportional  between  the  segments  of  the 
diameter. 

§  440.  Problem.  To  construct  the  mean  proportional  between  two 
given  lines. 

SIMILAR  POLYGONS 

§  442.  Problem.  Construct  a  polygon  similar  to  a  given  polygon 
having  given  a  side  corresponding  to  a  side  of  the  given  polygon. 

§  443.  Theorem.  The  corresponding  angles  of  two  similar  polygons 
are  equal. 


SYLLABUS   OF   PLANE  GEOMETRY  443 

§  444.  Theorem.  The  corresponding  sides  of  two  similar  polygons 
are  proportional. 

§  445.  Theorem.  The  perimeters  of  two  similar  polygons  are  in  the 
same  ratio  as  any  two  corresponding  sides. 

§  446.  Theorem.  The  areas  of  two  similar  polygons  are  in  the  same 
ratio  as  the  squares  of  any  two  corresponding  sides. 

PROPORTIONAL   LINES    CONNECTED   WITH   CIRCLES 

§  447.  Theorem.  If  two  chords  intersect  within  a  circle,  the  prod- 
uct of  the  segments  of  one  is  equal  to  the  product  of  the  segments  of  the 
other. 

§  450.  Theorem.  If  a  tangent  and  a  secant  are  drawn  to  a  circle 
from  an  external  point,  the  tangent  is  the  mean  proportional  between  the 
secant  and  its  external  segment. 

§  451.  Theorem.  If  a  tangent  and  a  secant  are  drawn  to  a  circle 
from  an  external  point,  the  square  of  the  tangent  is  equal  to  the  product 
of  the  secant  and  its  external  segment. 

§  452.  Theorem.  If  two  secants  are  drawn  to  a  circle  from  an  exter- 
nal point,  the  product  of  one  secant  and  its  external  segment  is  equal  to 
the  product  of  the  other  secant  and  its  external  segment.. 


Chapter  V.     Measurement  of  Circles 

§  458.     Problem.     To  inscribe  a  regular  hexagon  in  a  circle. 

§  459.  Theorem.  Any  equilateral  polygon  inscribed  in  a  circle  is  a 
regular  polygon. 

§  460.     Problem.     To  inscribe  a  square  in  a  circle. 

§  461.  Theorem.  If  a  circle  is  divided  into  any  number  of  equal 
parts,  the  chords  joining  the  successive  points  of  division  form  a  regular 
inscribed  polygon;  and  the  tangents  drawn  at  the  successive  points  of 
division  form  a  regular  circumscribed  polygon. 

§  462.  Theorem.  Tangents  drawn  to  a  circle  at  the  vertices  of  a 
regular  inscribed  polygon  form  a  regular  circumscribed  polygon  of  the 
same  number  of  sides. 

§  463.  Theorem.  Chords  drawn  from  the  middle  points  of  the  arcs 
subtended  by  the  sides  of  a  regular  inscribed  polygon  to  the  adjacent  ver- 
tices of  the  polygon  form  a  regular  inscribed  potygon  of  double  the  num- 
ber of  sides. 

§  464.  Theorem.  The  perimeter  of  a  regular  inscribed  polygon  is 
less  than  the  perimeter  of  a  regular  inscribed  polygon  of  double  the  num- 
ber of  sides,  and  less  than  the  circumscribed  circle. 


444  SOLID  GEOMETRY 

§  465.  Theorem.  The  area  of  a  regular  inscribed  polygon  is  less 
than  the  area  of  a  regular  inscribed  polygon  of  double  the  number  of  sides. 

§  466.  Theorem.  The  area  of  a  regular  polygon  is  less  than  the  area 
within  the  circumscribed  circle. 

POLYGONS   AND    CIRCLES 

§  469.     Problem.     To  divide  a  line  in  extreme  and  mean  ratio. 

§  470.     Problem.     To  construct  a  regular  decagon  in  a  given  circle. 

§  471.     Problem.     To  inscribe  a  regular  pentagon  in  a  circle. 

§  472.  Problem.  To  inscribe  in  a  given  circle  a  regular  pentadecagon, 
or  polygon  of  fifteen  sides. 

§  473.     Problem.     To  circumscribe  a  circle  about  any  regular  polygon. 

§  474.     Problem.     To  inscribe  a  circle  in  any  regular  polygon. 

§  476.  Theorem.  Two  regular  polygons  of  the  same  number  of  sides 
are  similar. 

§  477.  Theorem.  The  perimeters  of  two  similar  regular  polygons 
are  to  each  other  as  their  radii,  and  also  as  their  apothems. 

§  478.  Theorem.  The  area  of  a  regular  polygon  is  equal  to  half  the 
product  of  the  perimeter  by  the  apothem. 

§  479.  Theorem.  The  areas  of  two  regular  polygons  of  the  same 
number  of  sides  are  to  each  other  as  the  squares  of  their  radii,  and  also  as 
the  squares  of  their  apothems. 

.      MEASUREMENT   OF   THE   CIRCLE 

§  486.     Theorems  on  Limits. 

(1)  If  two  variables  that  approach  limits  are  equal  for  all  their  succes- 
sive values,  their  limits  are  equal. 

(2)  If  x  is  a  variable  having  the  limit  a,  and  c  is  a  constant  not  zero, 

.  x    a 
then  cx—>ca  and  — > -  as  x— >a. 
c    c 

§  488.  The  length  of  a  circle  is  the  common  limit  of  the  perimeters  of 
regular  inscribed  and  regular  circumscribed  polygons  as  the  number  of 
sides  is  indefinitely  doubled. 

§  489.  The  area  of  a  circle  is  the  common  limit  of  the  areas  of  regular 
inscribed  and  regular  circumscribed  polygons  as  the  number  of  sides  is 
indefinitely  doubled. 

§  490.  Theorem.  The  apothem  of  a  regular  inscribed  polygon 
approaches  as  a  limit  the  radius  of  the  circle,  as  the  number  of  sides  is 
indefinitely  doubled.  Also,  the  radius  of  a  regular  circumscribed  polygon 
approaches  as  a  limit  the  radius  of  the  circle,  as  the  number  of  sides  is 
indefinitely  doubled. 


SYLLABUS   OF   PLANE  GEOMETRY  445 

§  492.  Theorem.  Two  circumferences  have  the  same  ratio  as  their 
radii,  or  as  their  diameters. 

§  493.  Theorem.  The  ratio  of  a  circumference  to  its  diameter  is 
constant. 

§  495.  Theorem.  The  circumference  of  any  circle  is  equal  to  7Td,  or 
to  27Tr. 

§  496.  Theorem.  If  an  arc  s  is  intercepted  by  a  central  angle  of  u 

u 

degrees  the  length  of  the  arc  is TTd. 

360 

§  497.  Theorem.  The  area  of  a  circle  is  equal  to  one-half  the  product 
of  the  circumference  by  the  radius. 

§  498.     Theorem.     The  area  of  any  circle  is  equal  to  7Tr2. 

§  499.     Theorem.     The  area  of  any  circle  is  equal  to  \ird2. 

§  500.  Theorem.  The  areas  of  two  circles  are  in  the  same  ratio  as 
the  squares  of  their  radii  or  as  the  squares  of  their  diameters. 

§  508.  Theorem.  The  area  of  a  sector  is  equal  to  half  the  product 
of  its  radius  by  its  arc. 


INDEX 


Page 
Angles,  between  line  and  plane  294 

diedral 295 

face  .    . 302 

polyedral 302 

solid      302 

spherical 407 

triedral 303,413 

Circles,  of  sphere 384 

axis  of .  385 

great 384 

poles  of 385 

small 385 

Coilinear      .    .    . 280 

Concentric  spheres 392 

Cones,  area  of ■.  350 

axis  of .  344 

circumscribed      ....    350, 391 

definition  of 344 

frustum  of 353 

inscribed      350,391 

of  revolution 345 

similar      . 365 

spherical 403 

truncated 353 

vertex  angle  of 366 

volume  of 357 

Conical  surface 343 

Cube 326 

Cylinders 315 

area  of 315,319 

bases  of 315 

circular 317 

oblique 317 

of  revolution 317 

right 317 


Page 

Cylinders,  similar       336 

volume  of 321,331 

Cylindrical  surface 314 

Degree,  spherical    .    .    .    .    .    .421 

Devices  in  imaging 274 

Diedral  angle      295 

plane  angle  of     295 

Dimensions 273 

Directrix 314 

Distance,  between  parallel 

planes       290 

between  two  lines 300 

from  point  to  plane   ....  290 

on  spheres 386 

polar 386 

Dodecaedron 373 

Euler's  theorem      375 

Excess,  spherical     .    .    .    .    .   ..  421 

Face  angles     302 

Figures,  drawing    .    .  • 274 

Formulas 427 

Frustum,  area  of    354 

of  cone 353 

of  pyramid  .    .    .    .    .    .    .    .  353 

volume  of 363 

Generatrix 314 

Hexaedron 373 

Icosaedron 373 

Inclination 294 


447 


448 


INDEX 


Page 

Lines,  inclination  of 294 

parallel  to  planes 283 

perpendicular  to  planes     .    .  288 

projection  of 293 

relations  of      280 

tangent  to  sphere 382 

Loci      307 

Lime     ...... 398 

angle  of 398 

area  of 398 

Nappes 343 

Numbers 428 


Octaedron 


373 


Parallelpiped 326 

oblique 326 

rectangular      326 

right 326 

volume  of 327 

Planes,  coincide 280 

definition  of     .......  277 

determined      278 

how  represented 277 

intersecting 281 

parallel 285 

parallel  to  lines 283 

perpendicular 296 

perpendicular  to  lines    .    .    .  280 

relations  of      280 

tangent  to  cone  .    .    .  * .    .    .  349 
tangent  to  cylinder    .    .    .    .318 

tangent  to  sphere 383 

Point,  projection  of 293 

Polar  distance 386 

Polar  triangles 415 

Polyedral  angles 302 

central 409 

concave 302 

congruent ■  .    .  303 

convex 302 


Page 

Polyedral  angles,  parts  of     .    .  302 

symmetric 303, 410 

Polyedrons,  circumscribed     .    .  388 

concave 314 

convex 314 

definition  of     .......  313 

inscribed      388 

regular 373 

sections  of 313 

similar 377 

Prismatic  surface 314 

Prismatoid  formula 371 

Prismatoids,  definition  of      .    .  369 

volume  of    . 370 

Prisms      315 

area  of 315,319 

basis  of    .    .    : 315 

circumscribed      .    .    .    .    319, 357 

definition  of 315 

inscribed      ......    319,357 

oblique 317 

regular 317 

right 317 

similar 336 

truncated 325 

volume  of 331 

Projection 293 

Pyramidal  surface      343 

Pyramids,  areas  of 344 

circumscribed 350 

definition  of 344 

frustum  of 353 

inscribed      350 

regular 344 

similar 365 

spherical 423 

truncated 353 

volume  of .  357 

Quadrant 386 

Quadrilateral,  gauche  or  skew  .  283 


INDEX 


449 


Page 

Sector,  spherical 403 

Segment,  spherical     .    .    .    .    .  403 

Solids,  congruent    ......  324 

equivalent 324 

ratio  between      326 

Space  ideas 273 

Sphere,  and  straight  line   .    .    .381 

area  of 393 

circles  of 384 

circumscribed 391 

concentric 392 

definition  of 381 

distance  on      386 

great  circle  of 381 

inscribed  in  cone 391 

inscribed  in  cylinder      .    .    .391 

inscribed  in  polyedron  .    .    .  388 

small  circle  of 384 

tangent  to 392 

volume  of 400 

Spherical  angle 407 

Spherical  degree 421 

Spherical  cone 403 

Spherical  excess      421 

Spherical  polygon 408 

area  of 420 

congruent 410 

convex 409 


Page 

Spherical  polygon,  symmetric  .  410 

Spherical  pyramid 423 

Spherical  sector      403 

Spherical  segment      403 

Spherical  triangle 408 

area  of 421 

Spherical  wedge 405 

Surface,  conical      343 

cylindrical .314 

definition  of 277  . 

prismatic 314 

pyramidal 343 

Tangent  planes,  to  cone    .    .    .  349 

to  cylinder 318 

to  sphere 383 

Tetraedron      373 

Triangles,  polar      415 

spherical 408 

Triedral  angles   .....   303,  414 

isosceles 413 

Truncated  prisms 325 

Wedge,  spherical 405 

Zones 397 

area  of 397 

definition  of 397 


THIS  BOOK  IS  DUE  ON  THE  LAST  DATE 
STAMPED  BELOW 

EDUCATION  LIBRARY 

AN  INITIAL  FINE  OF  25  CENTS 

WILL  BE  ASSESSED  FOR  FAILURE  TO  RETURN 
THIS  BOOK  ON  THE  DATE  DUE.  THE  PENALTY 
WILL  INCREASE  TO  50  CENTS  ON  THE  FOURTH 
DAY     AND     TO     $1.00    ON     THE    SEVENTH     DAY 

OVERDUE. 

j  J  l  I:  o   Ioto 


1348 


LD  21-100w-12,'43  (8796s) 


^B  35956 


^o 


